YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Chapter 1 Radicals Lesson 1.1 Square roots and other Roots Square root The number a is a square root of b if a2 = b In algebraic form,
√ = a if and only if a2 = b.
All positive numbers have two square roots, one is positive and one is negative. The two square roots of 16 are 4 and -4, and the two square roots of 100 are 10 and -10. If b>0, the expression √ represents the principal ( or positive) square root of b. The principal square root of 0 is 0, that is the √ Example :√ If the integer is not a perfect square, then √
is an irrational number.
To find he principal square root of 2, enter 2 into a calculator and press the √ key, the approximate value of √ the symbol is read as “ is approximately equal to”.
Cube root The number a is a cube root of b if and only if a3 = b The cube root of b is denoted by √ . Example: √
, because 23 = 8
Root For any real number a and b, and any integer n>1, if an = b, then a is an nth root of b. in symbols, a = √ if an only if an = b.
Practice exercise 1.1 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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A. Find the indicated roots 1. √ 2. √ 3. √ 4. √ 5. √ B. Solve In a right triangle, the square of the hypotenuse c is equal to the sum of the squares of the sides a and b, c2 = a2 + b2. Find c if a = 7 cm and b = 24 cm. Lesson 1.2 Rational Exponents For any integer n>1 and any real number b for which √ Example: a. =√ =3 b. = √ =4
is defined,
= √ .
Writing the equivalent exponential expression of a radical expression 1. Write the radicand as the base of the exponential expression. 2. The reciprocal of the index becomes the exponent. We can extend the definition of to include rational exponents with a numerator other than 1. If m and n are positive integers with no common factor except 1, then for all real numbers b for which is defined. Power law of exponents
=
= √
) or √
Note : when is written as a radical, n is the index of the radical and m is the power to which the radical is raised. Example:
=( √
) =
=8
Negative Rational Exponents If m and n are positive integers with no common factor other than 1, and b then = , for all real numbers for which is defined
,
Example: YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Evaluate each expression. a. = = = √
Properties of rational exponents ( ) = = X0 = 1 = Example: a.
=
=
= 43 = 64
Answer the following: 1. Write each in radical form a. b. c. d. 2. Write in exponential form a. √ b. √ c. √ d. √ 3. Evaluate a. ( ) √
b. 811.5 c. d. (√
)
Lesson 1.3 Properties of Radicals The Product Rule for Radicals If a and b represent nonnegative real numbers, then √ √ =√ . The product of two square roots is the square roots of the product of the radicands. The square root of a product is equal to the product of the square roots of its factors. Proof: we assume that a and b are zero or positive constants. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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√
√ = = =√
Example: =√ =√ √ √ Rule Simplified form of a Square Root Radical Condition 1. Other than 1, the radicand has no perfect square factors. Condition 2. No fractions appear in the radicand. Condition 3. No radical appears in the denominator of a fraction. The product rule for radicals can be used to simplify radical expressions. Simplified square root radical expressions contain no factors that are perfect squares under the radical sign. We can use the product rule to simplify a radicand that is not a perfect square but has a perfect square factor, such as √ . the greatest square factor of 12 is 4, so we write 12 as 4 . Thus, √ = √ =√ √ = 2√ Because 3 has no perfect factor other than 1, the expression √ is the simplified form of √ . Simplifying square root radicals when radicands has a perfect square factor 1. Write the radicand as a product of the greatest perfect factor and another factor. 2. Use the product rule for radicals to write the expression as a product of two radicals with the perfect square factor as one radicand. 3. Evaluate the radical with the perfect square radicand. Example: a. Simplify √
, where x >0
Solution: factor 75x3 into two factors, one of which is the greatest perfect square that divides 75x3. Since 25 is the greatest perfect square that divides 75, and x2 is the greatest perfect square that divides x3, then the greatest perfect square that divides 75x3 is 25x2. Use the product rule for radicals to simplify √ =√ =√ √ = √ b. √ √ = √
=4
The product rule for higher-order Roots For real numbers a and b and any integer n>1 if n is even, then a and b must be nonnegative. √ √ = √ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Simplify: a. √
=√
= √ . √ = √ √ = 2√
The quotient rule for radicals If a and b represent nonnegative real numbers where b
, then
√
√ =√ . The square root of a quotient is equal to the quotient of the square roots of the numerator and denominator. Proof: we assume that a is 0 or positive and b is positive. √ =( ) =
=
√ √
Example: a.
√
=√
√
= 5x2 . 2
=√
Answer the following: 1. Simplify each quotient a. b. c. d. e.
√ √ √ √ √ √ √ √ √
Lesson 1.4 Addition and Subtraction of Radicals Radicals are like radicals when they have the same index and the same radicand. Since 3√ and √ contain like radicals, we can combine them. Thus, 3√ + √ =
√ = 4√
Similarly, 4 √
√
=
√ =
√
Rule of addition and subtraction of like radicals √ + √ =
√
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√
√ =
√
Example: add or subtract as indicated. a.
√
+ √ + √ √
b. √ + √ = √ +
√ √ √
√
√
√
=√ +
√ √ = √
=
√
√ =
√
√ =
√
√
c. Find the perimeter of the triangle with the given dimensions: a= 4√ b=4√ c=5√ P = (4√ + 4√ + 5√ = (4√ + 4√ + 5√ = (8√ + 12√ + 5√ = 25√
Practice: 1. √ 2. √ 3. √ 4. √
√ √ √ √
√ √
√
√
√
Lesson 1.5 Multiplication and Division of Polynomials containing Radicals The Quotient Rule for Radicals If a and b represent nonnegative real numbers where b
0, then
√ √
=√ .
The square root of a quotient is equal to the quotient of the square roots of the numerator and denominator. √ =( )
Definition of as exponent
( ) =
Product property of Exponents
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=
√
Definition of as exponent
√
Example: √ √
=√
quotient rule for radicals
=√ = 3x Simplified square root radicals: No radical appears in the denominator of a fraction. The expression √
is not simplified form because the radicand contains a fraction. This
√
can be written as because of the quotient rule for radicals. √ To eliminate the radical in the denominator, we will perform an operation that will result in a perfect square radicand in the denominator. Because
√ √
= 1 and √ . √ = 9, we
multiply both numerator and denominator by √ . Thus, √
√
√
√
√
. The process is called rationalizing the denominator because the denominator is change into a rational number. √
√
√
√
Multiplying a square root by itself √ √ = a where a is Example: a. √ = b.
√
=
√
√
√
√
√ √
√
√
√
=
= 2√
It is best to simplify the numerator and denominator before rationalizing the denominator. Example: Rationalize the denominator to simplify each radical a. √
=
√ √
=
√ √
√
=
√
√
√
√
=
√
=
√
A radical is in simplified form when the index of the radical is in its lowest possible form. Hence, if the index of a radical and the exponent of the radicand have a common factor, the expression can be written with a lower index. We call this process reducing the index.
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Example: a.
√
b. √
= =√
=√ =3
= =
=
=√
Practice: Simplify each radical. a. √ b.
√
c. √ d. √ e. √
Multiplying Radical with different Indices To find the product of radical when the indices are different but the radicands are the same, we follow these steps: Rules: Product of radicals with different indices 1. Transform the radicals to powers with rational exponents. 2. Multiply the powers by applying the rule xm xn = xm + n 3. Rewrite the product as a single radical 4. Simplify the resulting radical, if necessary. Product of radicals with different Indices and Different Radicands 1. Transform the radicals to powers with rational exponents. 2. Change the rational exponents into similar fractions. 3. Rewrite the product as a single radical. 4. Simplify the resulting radical, if necessary. Example Multiply and simplify = =√ √ √ =
=√
To multiply radical expressions with more than one term, use the distributive property to remove parentheses and combine like terms.
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Example: Multiply √ (√
)=√
√ √ =√ √ = 6x + 4√
Special Case Products (√
√ ) = (√ ) + 2√ √ + (√ ) = a + 2√
(√
√ ) = a - 2√
+b
+b
Example Square the radical expression (√
√ ) = (√ ) + 2√ √ + (√ ) = 2 + 2√ + 5 = 7 + 2√
Radical expressions can be conjugates. Conjugates involving radicals differ only in the sign joining the terms. For example, √ and √ are conjugates. The product of two conjugate binomials results in a difference of squares. (a + b) ( a – b) = a2 – b2 The same pattern occur when multiplying two conjugate radical expression Example Multiply (√
)(√
) = (√ ) - 32 =5–9 =-4
Division To divide radical expressions, we use the division rule for radicals. For example, to divide √ by √ , we proceed as follows. √ √
=√
=√
=5
Rationalizing the Denominator A radical expression is in simplified form as given in the third condition if no radical is in the denominator of a fraction. The process to remove a radical form from the denominator is called rationalizing the denominator. The nth root of a perfect nth power simplifies completely. Examples are: YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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√ √ √
=x =x =x
Hence, to rationalize a radical expression, use the multiplication property of radicals to create an nth root of an nth power. Rule: To rationalize the denominator of a radical expression, multiply the numerator and denominator by an expression that will give an nth root of an nth power in the denominator. Example Simplify the radical expression √
Solution √
= =
√ √
√ √
Practice I. Find the product 1. 5√ √ 2. 6√ √ II. Divide and simplify 1. 8√ √ 2. 32√ √ Lesson 1.6 Solving Radical Equations
Radical equation An equation containing at least one radical expression whose radicand has a variable is called a radical equation Extraneous Solution An apparent solution that does not solve the given equation Solving radical equations YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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To solve radical equation, follow the steps below. 1. Isolate the radical term. 2. Square both sides of the equation. 3. If all radicals have been eliminated, then solve. If a radical term remains, then isolate that radical term and squatre both sides again. 4. Check the solution. Example √ +1=5 √ =4 (√ ) = m(4)2 X = 16 Check: √ +1=5 4+1=5 5=5
The next examples require us to use the squaring property twice in order to eliminate the radical. Example -1=√ -2 √ =√ –1 √ X–3=x–2√ +1 -3 = -2 √ + 1 -4 = -2√ 2=√ 4=x
add 1 on both sides square both sides
Checking: -1=√ -2 √ -1=√ –2 √ √ -1=2–2 1 -1 = 2 – 2 0=0
The Pythagorean Theorem If the length of the hypotenuse of a right triangle is c and the lengths of the two legs are a and b, then c2 = a2 + b2 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Since the lengths of the sides of a triangle are positive numbers, we can use the square Root Property of Equality and the Pythagorean Theorem to find the length of the unknown side of a right triangle given the lengths of the other two sides. Example The sides of a square are 7 feet long each. Find the length of the diagonal. Solution: Draw the figure and let d be the length of the diagonal of the square. Use the Pythagorean Theorem to find the diagonal. 7 7
d2 = 49 + 49 d2 = 98 d= √ d= √ d = 7√ d = 7√ Practice Solve and check 1. √ = 18 2. √ = 35 3. √ Solve the problems 1. Find the length of the side of a square whose area is 50 square centimeters. 2. Find the length of the side of a square whose diagonal is 8 centimeters. 3. Find the length of one side of a cube whose volume is 54 cubic feet.
Chapter 2 Quadratic Equations and Functions Lesson 2.1 Solving quadratic equations by factoring Steps to be followed when factoring is applied in solving quadratic equations 1. Transpose all terms on the left side of the equation if necessary. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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2. 3. 4. 5. 6.
Combine similar terms Factor the left side of the equation Equate each factor to zero Solve the equation in step 4 Check each root by substituting in the original equation.
Some quadratic equations may be solved by factoring and by using the principle of zero property products. If ab = 0, then a = 0 or b = 0 Example: solve the quadratic equation 3x2 – 8x = 0 Solution: 3x2 – 8x = 0 x ( 3x – 8) = 0 x = 0 or 3x – 8 = 0 3x = 8 x= Practice I. 1. 2. 3. II.
Solve for x
- 144 = 0 Solve by factoring 1. 2. - 22x = 21 3.
Lesson 2.2 Solving Quadratic Equation by Completing the Square Completing the square In an expression of the form x2 + bx or x2 – bx, add the constant term ( ) to complete the square. Example YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Complete the square and then write each completed square in factored form. X2 + 14x Solution To complete the square add ( ) or 49. Completed square: x2 + 14x + 49 Factored form: ( x + 7)2
If the quadratic equation ax2 + bx + c = 0 has a leading coefficient of 1, that is a = 1, it is easy to solve by completing the square. In addition the equation should be written in the form x2 + bx = c. for example in the quadratic equation x2 + 24x = 36 the left side of the equation is x2 + 24x is an incomplete the square. We must complete the square by adding ( ) or 144. Because of the addition property of equality we must add 144 to both sides of the equation. Thus, X2 + 24x + 144 = 36 + 144 ( x+ 12)2 = 180 X + 12 = √ X = -12
√
X = -12
√
There are two solutions because of the Thus, x = -12 + 6√
sign.
or x = -12 - 6√
If a quadratic equation ax2 + bx + c = 0 has a leading coefficient other than 1, we can make the lead coefficient equal to 1 by dividing both sides of the equation by a. for the quadratic equation, 4x2 – 20x = 7, dividing both sides of the equation by 4 gives x2 a coefficient of 1.
X2 – 5x =
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We can solve by completing the square X2 – 5x +
= +
(
) =
(
) =8
X=
√
X=
√
X=
complete the square by adding
both sides
factor the left side expression. Add on the right
√
The steps in solving quadratic equation by completing the square 1. Transpose all terms containing the unknown to the left side of the equation and the constant term to the right side, if necessary. 2. Divide each term of the equation by the numerical coefficient of the x2 term, if necessary. This will change the equation to the form x2 + bx = c. 3. Divide the coefficient of x by 2, square it, and then add the result to both sides of the equation. 4. Factor the left side of the equation. This is perfect square trinomial. Simplify the right side. 5. Take the square root of the expression on both sides of the equation. Write the sign before the square root at the right side. 6. Equate the square root on the left side expression to the positive square root on the right side in step 5. Solve for the first root and then find the second root of the equation. 7. Equate the square root of the left side expression to the negative square root of the right side in step 5. Solve for the second root. 8. Check each root by substituting it to the original equation.
Practice I.
Given the number that must be added on both sides of each equation to complete the square. 1. X2 + 8x = 6 2. X2 – 7x = 8 3. 4x2 + 16x = 8
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II.
Solve each by completing the square. 1. X2 + 6x = 16 2. X2 + 4x = 45
Lesson 2.3 Solving Quadratic Equations by using the Quadratic Formula The quadratic formula Solving quadratic equations by using the quadratic formula The quadratic formula For real numbers a, b, and c with a + c = 0 is given by the quadratic formula
of the quadratic equation ax2 + bx
√
Use the quadratic formula to solve the equation 2x2 – 5x + 2 = 0, The quadratic equation is in standard form with a = 2, b = -5, c = 2. Substitute these values into the formula and simplify. √
√
√
√
=2
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= Procedure in solving quadratic equations using quadratic formula 1. 2. 3. 4. 5. 6.
Write the equation in standard form ( zero on one side of the equation) List the numerical values of the coefficients a, b, and c. Write the quadratic formula. Substitute the numerical values for a, b, and c. Simplify and get the exact solution Check the solution.
A quadratic equation can also have one solution or no real number solution Example: Solve the equation x2 + 3x + 6 = 0 √
√
√
√
Practice: Solve each using the quadratic formula 1. 2. 3. 4. 5.
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Lesson 2.4 Complex Numbers Imaginary Number i The imaginary number I has the following properties; I=√
and
i2 = -1
Definition of √ For any positive real number n, √ Because the definition of √
=i√
is for positive real number n, the radicand –n is negative.
Example Write the number in terms of i √
=√
=√
√
=i = Solve x2 = -4 =
√ √
Show that √ is square root of -5 Solution: multiply √
√ =
√ = i2
√ = -1
= -5
Rewriting Imaginary Numbers To write imaginary number√
, where n > 0 in terms of i:
1. Separate the radical into two factors; √ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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2. Replace √ 3. Simplify √
with i
Complex Number A complex number can be written in the standard form a + bi where a and b are real numbers. The number a is called the real part, and the number b is called the imaginary part. The following are some examples of complex numbers in standard form; 7 – 3i
3 + 4i
4.5 – 2i√
If a = 0, then the complex number is purely imaginary number, such as -8i
and
2i√
If b =0, then the complex number is a real number. Equality of Complex Numbers Two complex numbers a + bi and c + di are equal, if and only if a = c and b = d Example: 2 + 5i = √
+
i because 2 = √ and 5 =
We can perform arithmetic operations with complex numbers. We treat the complex numbers just like polynomials, where I is like a variable. Addition and subtraction of complex numbers Complex numbers are added and subtracted as if they were binomials: (a + bi) + ( c + di) = (a + c) + ( b + d) i Example: add or subtract (12 – 9i) + ( 8 + 3i) = 20 – 6i To multiply a complex number by an imaginary number, we use the distributive property to remove the parentheses and then simplify. For example, -4i (7 – 6i) = -4i (7) – (-4i)(6i) = -28i + 24i2
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= -28i - 24 = -24 – 28i Multiplication of complex numbers Complex numbers are multiplied as if they were binomials: (a + bi)(c + di) = (ac – cd)( ad + bc)i Example: Multiply ( 4+ 5i)(2 – 3i) = 8 – 12i + 10i – 15i2 = 8 – 2i + 15 = 23 – 2i (7 + i)( 6 + 5i) = 42 + 35i + 6i + 5i2 = 42 + 41i – 5 = 37 + 41i
Complex conjugates The complex numbers a + bi and a – bi are called complex conjugates of each other 5 + 2i and 2 – 2i are complex conjugates. 9 – 3i and 9 + 3i are complex conjugates Example: find the product of 8 + I and its complex conjugate. Solution: the complex conjugate of 8 + I is 8 – i. We find the product as follows: (8+i)(8-i) = 64 – 8i + 8i – i2 = 64 – i2 = 64 – (-1) = 65 In general, the product of a complex number a + bi and its conjugate a – bi is a real number a2 + b2. (a + bi)(a –bi) = a2 – abi + abi – b2i2 = a2 – b2 – (-1) = a2 + b2 To transform complex numbers such as
in a + bi form, we rationalized
their denominators. Example:
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=
=
=
We rationalized the denominator by multiplying the numerator and denominator by the complex conjugate of the denominator and simplify. Practice: 1. Express in terms of i a. √ b. √ c. √ d. 4√ e. √ 2. Simplify each sum or difference a. (9 + 5i)+(8 + 9i) b. (7 + 6i) + ( 11 + 10i) c. (-2 + 8i) + ( 4 – 3i) d. (12 + 17i) – (13 – i) e. (-15 - √ -(3-√ 3. Simplify each product a. (4 + 2i)(7 – 4i) b. √ (√ ) c. i) d. i)(-6 + 2i) e. ( √ )( √ ) 4. Write the conjugate of each complex number. a. 8 + 3i b. -9 + 4i c. 7 + 7i d. -12 – 9i e. 11 + 11i 5. Simplify each quotient a. b. c.
√ √
d. e. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Lesson 2.5 Roots and Coefficients of quadratic equations Discriminant the radicand b2 – 4ac in the quadratic formula is called the disciminant. We use the discriminant to determine the number of real solutions of a quadratic equation. It also tells whether those solutions are either rational or irrational numbers. Using the discriminant Given a quadratic equation in the form ax2 + bx + c = 0, where a, b, and c are real numbers and a we can determine the number and type of solutions of a quadratic equation by evaluating the discriminant b2 – 4ac. 1. If b2 – 4ac > 0, the equation ha b2 – 4ac has two real solutions. Both will be a rational b2 – 4acif the discriminant is a perfect square or irrational, if otherwise. 2. if b2 – 4ac = 0, the equation has only one solution which will be a rational number. 3. if b2 – 4ac < 0, the equation has no real number solution. Example 1 Use the discriminants to determine the number of solutions in 16x2 – 8x + 1 = 0 Solution: Evaluate the discriminant b2 – 4ac b2 – 4ac = (-8)2 – 4(16)(1) = 64 – 64 = 0 Because the discriminant is 0, the equation has only one solution x2 + 2x + 3 = 0 b2 – 4ac = (2)2 – 4 (1)(3) = 4 – 12 = -8 because the discriminant is negative, the equation has no real number solution 2x2 – 3x = 4 2x2 – 3x – 4 = 0 b2 – 4ac = (-3)2 – 4 (2)(-4) = 9 + 32 = 41 Because the discriminant is positive and not a perfect square, the equation has two irrational solutions. Example 2. For the equation 3x2 – 8x + c = 0 For the equation to have one irrational number solution, the discriminant b 2 – 4ac must be equal to 0. b2 – 4ac = 0 (-8)2 – 4(3)c = 0 64 – 12c = 0 or
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To check whether will make the equation 3x2 – 8x + c = 0 to have one rational number solution, we have to substitute this value for c in the equation. 3x2 – 8x + = 0 9x2 – 24 x + 16 = 0 (3x – 4)2 = 0 3x = 4 x= Example: write the quadratic equation given a pair of solutions 7, -2 Solution: X = 7 or x = -2 Reverse the factoring method using solution 7 and -12. x -7 = 0 or x+2=0 (x -7) (x + 2) = 0 x2 – 5x – 14 = 0 Practice: 1. State whether the roots of the following equations are real and unequal, real and equal or imaginary and unequal: a. x2 + 6x + 9 = 0 b. 5x2 – x = x – 4 c. + = x – 4 d. x ( x – 5) = 4(5 – x) e. x2 + 7x + 1 = 0 Lesson 2.6 Systems of Equations Involving Quadratic Equations A quadratic system involves at least one quadratic equation. A quadratic system with at least one linear equation is called a quadratic-linear system. A quadratic system with two quadratic equations is called quadratic –quadratic system.
A system of equations which consists of a linear equation and quadratic equation has two roots while a system of equations which consists of two quadratic equations usually have four roots. As with the linear the linear system, a quadratic system can be solved by; 1. Graphing, 2. Substitution, or YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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3. Linear combination. Example 1 the system can be solved by substitution ab = 36
equation 1
a2 + b2 = 97
equation 2
Change the equation to b-form: ab = 36 b=
equation 3
Substitute
for b in equation 2.
( )
Let a2 = u. thus, we have
or
Solve for a: if u = 81 a2 = u a2 = 81 a= √ a= Solve for b: If a = 9 if a = -9 ab = 36 ab = 36 9b = 36 -9b = 36 b=4 b = -4
if u = 16 a2 = u a2 = 16 a= √ a= if a = 4 ab = 36 4b = 36 b=9
if a = -4 ab = 36 -4b = 36 b = -4
there is no fixed procedure in solving a system of equations involving quadratic equations but oftentimes, the target is to find the values of any of the following expressions: x + y, x – y, and xy from which the value of x and y can be obtained.
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Example 2 Solve the system x + y = 10 and xy = 24 x + y = 10
Equation 1
xy = 24
Equation 2
Squaring equation 1, we have Equation 3 Multiplying equation 2 by 4, the result is 4xy = 96
Equation 4
Subtracting equation 4 from equation 3, the difference is X2 – 2xy + y2 = 4
equation 5
Factoring and getting the square root of equation 5, we have √
=√
x–y=
equation 6
Adding equations 1 and 6, we have (1)
x + y = 10 x–y=2 2x
= 12 x=6
or
x + y = 10 x – y = -2 2x
=8 x=4
To solve for y, we have If x = 6, then 6 + y = 10 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Thus,
y = 10 – 6 Y=4
If x = 4, then 4 + y = 10 Thus,
y = 10 – 4 Y=6
Therefore, the two solutions are (6,4) or (4,6) Example 3 Solve the given systems of equations. 4x2 + y2 = 52
equation 1
2x + y = 10
equation 2
Solution: Use the substitution method to solve by rewriting Equation 2. Thus, y = 10 – 2x 4x2 + y2 = 52 4x2 + (10 – 2x)2 = 52 4x2 + 100 – 40x + 4x2 = 52 8x2 – 40x + 48 = 0 X2 – 5x + 6 = 0 (x – 3)(x – 2) = 0 X – 3 = 0 or x – 2= 0 X = 3 or x = 2 Now solve for y: If x = 3 Y = 10 – 2 (3) YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Y=4 If x = 2, then Y = 10 – 2(2) Y=6 The solutions of the system of equations are (3,4) and (2,6) Example: Solve the given system of equations Y + 12x = 50 6x2 + y2 = 100 Y + 12x = 50 eq.1 6x2 + y2 = 100
eq.2
Solve equation 1 for y Y = 50 – 12x eq. 3 Substitute 50 – 12x in place of y in eq.2 6x2 + ( 50 – 12x)2 = 100 6x2 + 2500 – 1200x + 144x2 = 100 150x2 – 1200x + 2400 = 0 X2 – 8x + 16 = 0 (x – 4)2 = 0 X–4=0 X=4 When x = 4 6(4)2 + y2 = 100 Y2 = 4 Y= YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Thus, there are two possible solutions: (4, -2) and (4, 2). The point ( 4,2) satisfies the equation y + 12x = 50 but the point (4,-2) does not. The solution (4,-2) is an extraneous solution. Hence, the system has only one solution. Practice: Solve the following systems of equations 1. x2 – y2 = 33 xy = 28
2. x + y = 16 xy = 55
3. x – y = 3 x2 + y2 = 185
Lesson 2.7 Quadratic Inequalities A quadratic inequality is an inequality of the form ax2 + bx + c < 0, where a, b, and c are real numbers with a 0. The inequality symbols >, , may also be used. If a quadratic inequality is factorable, it can be solved using a sign graph. A sign graph shows where each factor is positive, negative, or zero.
Example: solve the inequality x2 + x – 6 > 0 Solution: The left hand side can be factored, hence we can write the inequality as ( x + 3) ( x – 2 ) > 0
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The inequality says that the product of x + 3 and x – 2 is positive. If both factors are positive or both are negative, the product is positive. To analyze the signs of each factor, we can make a sign graph as follows. Consider the possible values of the factor x + 3: value x+3=0 x+3>0 x+3<0
where If x = -3 If x > -3 If x < -3
On the number line Put 0 above -3 Put + sign to the right of -3 Put – sign to the left of -3
Now, consider the possible values of the factor x – 2: value x–2=0 x–2 >0 x–2 <0
where If x = 2 If x > 2 If x < 2
On the number line Put 0 above 2 Put + sign to the right of 2 Put – sign to the left of 2
The information for the factors x + 3 and x – 2 are shown on the sign graph at the right. We can see from the sign graph that the product is positive if x < -3 and also positive if x > 2. The solution set of the inequality x2 + x – 6 > 0
-5
l
l
l
l
l
l
l
l
l
-4
-3
-2
-1
0
1
2
3
4
5
Note that -3 and 2 are not included in the graph because for those values of x, the product is zero. In symbols, the solution set is (
Strategy for solving a quadratic inequality with a sign graph 1. 2. 3. 4.
Write the inequality with 0 on the right side. Factor the quadratic trinomial. Prepare a sign graph showing where each factor is positive, negative, or zero. Use the rules for multiplying signed numbers to determine which regions satisfy the original inequality
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Example Solve x2 + 18
and graph the solution set.
Solution: Method 1 Step 1. Write the inequality with 0 on the right x2 + 18 x2 – 9x + 18
subtract 9x from both sides
Step 2. Factor the quadratic trinomial x2 – 9x + 18 ( x – 6)( x – 3)
Step 3. Examine the signs of each factor x – 6 = 0 if x = 6 x – 6 > 0 if x > 6 x – 6 < 0 if x < 6 x – 3 = 0 if x = 3 x – 3 > 0 if x > 3 x – 3 < 0 if x < 3
Step 4. The product of x – 6 and x – 3 is positive if both factors are both positive or both negative and the product is 0 if one of the factors is 0.
Strategy for solving a rational Inequality with a sign graph 1. 2. 3. 4.
Rewrite the inequality with 0 on the right hand side. Get an equivalent inequality using addition or subtraction Factor the numerator and denominator, if possible. Prepare a sign graph.
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5. Use the rules for multiplying and dividing signed numbers to determine the region that satisfies the original inequality.
Example: Solve and graph the inequality Solution:
Since x – 3 and 3 – x are additive inverses, we obtain the LCD by multiplying the numerator and denominator of one of the rational expressions by -1. We choose the simplified rational expression.
Using the rule for dividing signed numbers and he sign graph, we can identify whether the quotient is positive, negative or zero. The solution set is (
)
. Note that
3 is not in the solution set because the quotient is undefined if x = 3
Practice: A. Determine the ordered pair is a solution of the given quadratic inequality. 1. x2 –x – 12 > 0 , ( -3,4) 2. 3. B. Solve and graph the solution on a number line 1. 2. 3. 4. 5.
(x+5)(x – 4) > 0 (x – 2 ) (x + 3) > 0 (x + 3)(x – 5) < 0 (x + 2) ( x + 5)
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Chapter 3 Variation
Lesson 3.1 Direct Variation The value of y varies directly with the value of x if there exists a nonzero constant k such that y = kx or, equivalent,
=k
The constant k is called the constant of variation. If y varies directly as x and we know the corresponding values of x and y, then we can find the proportionality constant k. Once k is known, we can write the equation of variation and use it to determine other values. Also, if we know that y varies directly as x and one set of values, we can use the proportion to find the other sets of corresponding values. y1 = kx1 =k y2 = kx2 =k Therefore,
=
Example: The table shows the number of hours (t) travelled and the distance (y) covered by a car. No. of hours travelled (t) 1 2 3 4 5 Distance in km covered (y) 40 80 120 160 200 1. Show that t varies directly as y. 2. Draw a graph of y against t. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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3. Write the equation showing the relationships between t and y. 4. Find y when t = 1.5 and t = 4.25 5. Find t when y = 140 and y = 260 Solution: T 1 2 3 4 5 Y 40 80 120 160 200 40 40 40 40 40
Because
= 40 is a constant, y varies directly as t.
Because
= 40, the equation relating t and y is y = 40t
When t = 1.5 y = 40t = 40(1.5) = 60 When t = 4.25 y = 40(4.25) y = 170
when y = 140 140 = 40t t = 3.5 when y = 260 260 = 40t t = 6.5
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Another example A workerâ&#x20AC;&#x2122;s paycheck (P) varies directly as the number of hours (h) worked. For working 20 hours, the payment is â&#x201A;ą1000. Find the payment for 45 hours of work. When h = 20, P = 1000 1000 = k
20
k= k = 50 Thus, P = 50h
When h = 45 P = 50 = 2250 P=
Direct Variation as a Power The value of y varies directly as the nth power of x if there exist a non zero real number k such that y = kxn
One example involves the area of a circle. That is, A= Here, is the constant of variation, and the area varies directly as the square of the radius r.
Example: The distance a body falls from rest is directly proportional to the square of time it falls. (a) If an object falls 20 meters in 2 seconds, how far will it fall in 8 seconds? (b) If an object is dropped from a height 405 meters, find the time it takes to hit the ground. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Solution:
a. The distance d varies as the square of the time t Then, d = kt2 20 = k(2)2 20 = k4 k=5 The proportionality constant of variation equation is 5. Thus, d = 5t2 When t = 8, d = 5(8)2 = 320 The object will fall 320 meters in 8 seconds b. when d = 405 m 405 = 5t2 t2 = 81 t=9 Thus, it will take an object from a height of 405 meters in 9 seconds to hit the ground. Practice: 1. The electric current I (in amperes) in an electric circuit varies directly as the voltage V. When 24 volts is applied, the current is 8 amperes. Find the current when 12 volts is applied. 2. The horsepower produced by a stream engine varies directly as the number of revolutions per minute of its operation. If 1200 horsepower was produced when a machine was operated at 150 revolutions per minute, find the horsepower it will produce when it will operate at 170 revolutions per minute.
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3. The weight W of an object on Mars is directly proportional to its weight B on Earth. A man who weighs 47.5 kg on Mars. Find the weight of a man on Mars if he weighs 98 kg on Earth. 4. The amount of pollutants A entering the atmosphere in an area with a population P living in an area. If 25000 tons of pollutants enter the atmosphere in an area with a population of 17750, how many tons of pollutants enter the atmosphere in Manila with a population of 2 450 500? 5. The rate of a bus that rolls down on a hill is directly proportional to the length of time it rolls. At the end of 10 seconds, its rate is 30 mph. how long will it take the bus to have a rate of 42 mph? 60mph? 81 mph?
Lesson 3.2 Inverse Variation
The quantity y varies inversely with x if there exists a nonzero constant k such that y = or equivalently. xy = k Also, y varies inversely as the nth power of x if there exist a nonzero real number k such that y= A proportion can also be used in relation to indirect variation in solving problems where some quantities are known. The following proportions are two of several proportions that can be formed: x1y1 = k
x2y2 = k
x1y1 = x2y2 =
Example: Suppose a varies inversely with b. when b = 4 and a = , find a when b = 18 Solution: YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Since a varies inversely with b a= When b = 4 and a = = k=2 thus, a = when b = 18 a= Example: The law of the lever states that: “ to balance a given person seated on a seesaw, the distance d the other person is from the fulcrum is inversely proportional to that person’s weight w. “ Bongbong and Abet are trying to balance on a seesaw. Bongbong weighs 100lb and sits 4 feet from the fulcrum. Abet weighs 80 lb. how far from the fulcrum must Abet sit to balance the seesaw with Bongbong? Solution: Since d and w are in inverse variation, d= find k substituting Bongbong’s weight and distance into d = 4= k = 400 thus, the variation formula is d= evaluate when w = 80 lb. d= YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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d=5 Abet must sit 5 feet from the fulcrum Example: The number of hours t required to finish a certain job varies inversely as the number of persons N on the job. If 16 persons require 18 hours to finish the job, how long would it take 64 persons to finish the job? Solution: Because t and N are in inverse proportion t1N1 = t2N2 let t1 = 18, N1= 16 and N2 = 64 18(16) = t2(64) 188 = 64t2 t2 = 4.5
Practice: Solve the following 1. The bases of triangles having equal areas are inversely proportional to their altitudes. The base of a certain triangle is 24 cm and its altitude is 30 cm. find the base of a triangle whose altitude is 40. 2. The intensity, i, of light from a bulb is inversely proportional t the square of the distance, s, from the bulb. If i=90W/m2 when the distances s = 20, find the intensity at a distance of 10 m. 3. Boyleâ&#x20AC;&#x2122;s Law states that when gas is put under pressure, the volume is inversely proportional to the pressure. State this law as a proportion. 4. The weight of an object is inversely proportional to the square of its distance from the center of the earth. At sea level ( 6400 km from the center of the earth), a cosmonaut weighs 110 kg. Find his weigh when he is 210 km above the surface of the Earth. 5. The current I in amperes is inversely proportional to the resistance R in ohms. If the current is 30 amperes when the resistance is 3 ohms, what is the current when the resistance is 5 ohm? YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Lesson 3.3 Joint Variation and Combined Variation
Joint Variation Y varies jointly as x and z if there exists a nonzero constant k such that y = kyz , where x 0 and z 0
The area of a triangle depends on its base b and its altitude h by the formula: A = kbh 12 = k(6)(4) 12 = 24k k= the equation for the area of a triangle is A = bh, where A is the area, b is the base, h is the altitude, and is the constant variation. Example: Suppose y varies jointly as x and z and y = 32 when x = 8 and z = 2. Find y when x = 15 and z = 6
Solution: y = kxz 32 = k(8)(2) 32 = 16k k=2 Rewriting the equation y = kxz with k = 2 we have y = 2xz To find y when x = 15 and z = 6 we have YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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y = 2(15)(6) y = 180
Combined Variation If a situation is modeled by an equation of the form y= Where k is a nonzero constant, we say that y varies directly as x and inversely as z. The number k is called the variation constant.
Example: If y varies directly with square of x and inversely with z and x = 2, y = 1 and z = 10, find y when x = 4 and z = 5 Solution: y= 1= 10 = 4k k= The constant of variation k is
or
To find y when x = 4 and z = 5, replace x with 4 and z with 5. Thus, y= y= y= y=8 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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More Examples of Variation Statement y is directly proportional to the square root of x y varies directly as the square of x
Formula y=k√ y = kx2
y varies inversely as the square root of x
y=
y is inversely proportional to the cube of x
y=
y varies jointly as x and the square root of z
y = kx√
y varies directly with the cube of x and inversely with the square root of z
y=
√
√
Practice: 1. x varies directly as y and inversely as z. If x = 30 when y = 40 and z = 80, find x when y = 24 and z = 40. 2. c varies jointly as a and b. If c = 90 when a = 30 and b = 28, find c when a = 42 and b = 16. 3. t varies directly as m and inversely as the square of n. If t = 32 when m = 16 and n = 4, find t when m = 24 and n = 6 4. m varies jointly as n and p. If p = 18 when m = 144 and n = 16, find p when m = 24 and n = 16. 5. q varies jointly as g and the square of b. If q = 210 when g = 28 and q when g = 20 and b = 28.
Chapter 4 Quadrilaterals Lesson 4.1: Ways of Proving That Quadrilaterals Are Parallelograms Theorem 50 Each diagonal of parallelogram divides the parallelogram into two congruent triangles A B
D
C
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If If
ABCD is a parallelogram with diagonal ̅̅̅̅ , then
ABC
CDA.
ABCD is a parallelogram with diagonal ̅̅̅̅ then ABD
Theorem 51 Opposite sides of a parallelogram are congruent A
B
D
If ABCD is a parallelogram, then ̅̅̅̅
C
̅̅̅̅ and ̅̅̅̅
̅̅̅̅ .
Theorem 52 Opposite angles of a parallelogram are congruent
If ABCD is a parallelogram, then angle A Theorem 53 Any two consecutive angles of a parallelogram are supplementary. If ABCD is a parallelogram, then angle A and angle B are supplementary angles. Theorem 54 The diagonals of a parallelogram bisect each other. If ABCD is a parallelogram, then ̅̅̅̅
̅̅̅̅
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̅̅̅̅ Page 43
Theorem 55 If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.
If ̅̅̅̅
̅̅̅̅
̅̅̅̅
̅̅̅̅, then ABCD is a parallelogram.
Theorem 50 Each diagonal of parallelogram divides the parallelogram into two congruent triangles A
D If If
B
C
ABCD is a parallelogram with diagonal ̅̅̅̅ , then
ABC
CDA.
ABCD is a parallelogram with diagonal ̅̅̅̅ then ABD
Theorem 51 Opposite sides of a parallelogram are congruent A
D
If ABCD is a parallelogram, then ̅̅̅̅
B
C
̅̅̅̅ and ̅̅̅̅
̅̅̅̅ .
Theorem 52 Opposite angles of a parallelogram are congruent
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If ABCD is a parallelogram, then angle A
Theorem 53 Any two consecutive angles of a parallelogram are supplementary.
If ABCD is a parallelogram, then angle A and angle B are supplementary angles.
Theorem 54 The diagonals of a parallelogram bisect each other. If ABCD is a parallelogram, then ̅̅̅̅
̅̅̅̅
̅̅̅̅
̅̅̅̅
Theorem 55 If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.
If ̅̅̅̅
̅̅̅̅
̅̅̅̅
̅̅̅̅, then ABCD is a parallelogram.
Theorem 56 A quadrilateral is a parallelogram if both pairs of opposite sides are congruent
If ̅̅̅̅
̅̅̅̅ and ̅̅̅̅
̅̅̅̅ , then ABCD is a parallelogram.
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Theorem 57 A quadrilateral is a parallelogram if both pairs of opposite angles are congruent. If angle A
angle C and angle B
angle D then
ABCD is a parallelogram.
Theorem 58 A quadrilateral is a parallelogram if two sides are both parallel and congruent. If ̅̅̅̅
̅̅̅̅ and ̅̅̅̅
̅̅̅̅ , then
ABCD is a parallelogram.
Theorem 59 If an angle of a quadrilateral is supplementary to both consecutive angles, then the quadrilateral is a parallelogram. If angle A and angle B are supplementary angles and angle A and angle D are supplementary angles, then ABCD is a parallelogram.
Test I.
II.
Decide whether the information given is enough to determine that a quadrilateral is a parallelogram 1. Each diagonal bisects opposite angles 2. Both pairs of opposite sides are congruent 3. Two pairs of consecutive sides are congruent Determine whether the quadrilateral with the given vertices is a parallelogram. 1. K(5,6), I(9,0),T(8,-5), and E(3,-2) 2. T(4,4), U(9,3),N(9,7) and E(5,8) 3. R(-7,3),A(-3,2),I(0,-4) and N(-4,-3)
Lesson 4.2 Special Parallelograms
A square is an equiangular rhombus. A square is an equilateral rectangle. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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Theorem 61 The diagonal of a rhombus are perpendicular and bisect each other. If
ABCD is a rhombus, then ̅̅̅̅
̅̅̅̅
̅̅̅̅
̅̅̅̅.
Theorem 62 The diagonals of a square are congruent and are perpendicular. If
ABCD is a square, then ̅̅̅̅
̅̅̅̅
̅̅̅̅
̅̅̅̅.
Theorem 63 Each diagonal of a rhombus bisects two angles of the rhombus. If
ABCD is a rhombus, with diagonal ̅̅̅̅
̅̅̅̅ bisects ⦟B and ⦟D.
Theorem 64 If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. If
ABCD is a parallelogram, and ̅̅̅̅
̅̅̅̅
. then
ABCD is a rectangle.
Theorem 65 If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. If
ABCD is a parallelogram, with ̅̅̅̅
̅̅̅̅
ABCD is a rhombus.
Theorem 66 If the diagonals of a parallelogram are congruent and perpendicular, then the parallelogram is a square. If
ABCD is a parallelogram, and ̅̅̅̅
̅̅̅̅
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̅̅̅̅
ABCD is a square. Page 47
Theorem 67 If one diagonal of a parallelogram bisects two angles of the parallelogram, then the parallelogram is a rhombus. If
ABCD is a parallelogram with ̅̅̅̅ bisecting ⦟B and ⦟D, then
ABCD is a rhombus.
Squares have all the properties of rectangles and rhombuses. Because rectangles and rhombuses are both parallelograms, they have all the properties of a parallelogram To summarize the different characteristics of the special parallelograms In a rectangle 1. 2. 3. 4. 5. 6. 7.
Opposite sides are congruent Opposite sides are parallel Each diagonal separates the rectangle into two congruent triangles. Opposite angles are congruent. Consecutive angles are supplementary. All four right angles are right angles. Diagonals bisect each other and are congruent
In a rhombus 1. 2. 3. 4. 5. 6. 7.
All four sides are congruent. Opposite sides are parallel. Each diagonal separates the rhombus into two congruent triangles. Opposite angles are congruent. Consecutive angles are supplementary, Diagonals bisect each other and are perpendicular. Each diagonal bisects a pair of opposite angles.
In a square 1. 2. 3. 4. 5. 6.
All four sides are congruent. All angles are right angles. Each diagonal separates the square into two congruent triangles. Opposite angles are congruent and supplementary. Consecutive angles are supplementary and congruent. Diagonals bisect each other, are perpendicular and congruent.
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Test Name all the quadrilaterals – parallelogram, rhombus, rectangle, or square – with given properties. 1. 2. 3. 4. 5.
Diagonals are congruent. Diagonals bisect each other. Diagonals are perpendicular. All sides are congruent. One pair of opposite angles are congruent.
Lesson 4.3 Trapezoids and Kites A trapezoid is a convex quadrilateral with exactly one pair of parallel sides The bases of a trapezoid are the two parallel sides. The legs of a trapezoid are the two non-parallel sides. An isosceles trapezoid is a trapezoid with congruent legs. The median of a trapezoid is the segment that joins the midpoints of the legs. An altitude of a trapezoid is any segment from a point on one base perpendicular to the line containing the other base. Theorem 68 The median of a trapezoid is parallel to its bases, and its length is half the sum of the lengths of the bases.
Theorem 69 The base angles of an isosceles trapezoid are congruent. If
ABCD is an isosceles trapezoid, then ⦟D
⦟ .
Theorem 70 If the base angles of a trapezoid are congruent, then the trapezoid is isosceles. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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If ⦟D
⦟C in trapezoid ABCD, then
ABCD is an isosceles trapezoid
Theorem 71 The diagonals of an isosceles trapezoid are congruent. If
ABCD is an isosceles trapezoid, then ̅̅̅̅
̅̅̅̅.
Theorem 72 If the diagonals of a trapezoid are congruent, then the trapezoid is isosceles. If in trapezoid ABCD, ̅̅̅̅
̅̅̅̅, then trapezoid ABCD is isosceles.
Kite Another special quadrilateral that is not a parallelogram is a kite. Like a rhombus, the diagonals of a kite are perpendicular.
A kite is a quadrilateral with two pairs of adjacent sides congruent and no opposite sides congruent. The common vertices of the congruent sides of kite are the ends of the kite. The line containing the ends of a kite is a symmetry line for the kite.
Theorem 73 The diagonals of a kite are perpendicular If ABCD is a kite, then ̅̅̅̅
̅̅̅̅.
Every square is a rhombus. Every square is a rectangle.
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Test Decide whether each statement is true or false. 1. 2. 3. 4.
The diagonals of a trapezoid are congruent. The parallel sides of an isosceles trapezoid are congruent. The opposite angles of a kite can be supplementary. The length of the median of a trapezoid is one-half the sum of the lengths of the bases.
Chapter 5 Similarities Lesson 5.1 Ratio and Proportion
Properties of proportion Manipulation =
( )
Reasons Fraction form of proportion MPE
Equations =
Examples using a = 2, b =3, c =4, d=6 = (the right side can be reduced to ) = ( both sides can be reduce to
=
) ( )
*( )
+
MPE
=
Reciprocal of both sides MPE
=
= (both sides can be reduced to 2) ( the right side can be reduced to ) 2 to 12)
( both sides are equal
APE
=
(both reduce by )
ides
can
be
SPE
=
(both reduce by )
ides
can
be
Example
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Solve for x = 15x = 60 x=4
Another example Find the fourth term in a proportion if the first three terms are 12,17 and 36, respectively, Solution: Let x be the fourth term of the proportion. So, = Solve for x 112x = 612 X = 51
Practice A. Find the ratio of x and y 1. 4x = 5y 2. mx = ny 3. xa + xb = yc + yd B. find the fourth number in each proportion if the following set of numbers are the first three numbers of the proportion. 1. 8, 12, 32 2. 15, 60, 4 3. 11,23,33
Lesson 5.2 Similar Triangles and Polygons
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If two triangles are similar, then a. The corresponding angles are congruent, and b. The corresponding sides are proportional. If
, then ⦟ A
⦟
⦟
⦟ , ⦟C
⦟
and
The common ratio of the lengths of the corresponding sides of similar polygons is also called the scale factor of the similar polygons. The scale factor of the similar quadrilaterals is 2:1 since both figures have the same unit of measure. The scale factor is independent of the unit of measure. One important application of similar polygons is scale drawing. Blueprints and road maps are examples of scale drawings. The unit of measure in a scale drawing is different from its actual unit of measure. For example, a blueprint may be measured in centimeters while the room it represents may be in meters. To interpret scale drawings you must know the scale ratio, including its unit of measure.
Theorem 74 Perimeters of Similar Polygons If two polygons are similar, then the ratio of their perimeters is equal to the ratio of their corresponding sides. If
, then
and
Example In the figure, M
C A
N
T
G
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a. Find the value of x. b. Find the ratio of the perimeter of
to the perimeter of
Solution: a. Because
, Replace CA with 8, MN with 16, CT with 10, and MG
with x. Apply the cross-product property of Proportion. Divide each side by 8. b. Using the perimeter of Similar Polygons Theorem, the ratio of the perimeter of to the perimeter of
is
Practice 1. Given: a. b. c. d.
.
Name all pairs of corresponding sides. Name all pairs of corresponding angles. For the ratio of similarity or scale factor is _____. If HS = 9, then MU = _____
Lesson 5.3 Proving Similar Triangles
Postulate 21 The AAA Similarity Postulate If three angles of one triangle are congruent to three angles of another triangle, then the two triangles are similar. If ⦟A
⦟
⦟
⦟
⦟
⦟
B
E D
A
F
C
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Theorem 75 The AA Similarity Theorem If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. If ⦟A
⦟
⦟
⦟ , then
If two angles of one triangle are congruent respectively to the corresponding angles of another triangle, then the third pair must also be congruent.
Corollary 75 ̅̅̅̅
If in third side ̅̅̅̅ and then
̅̅̅̅ at P and Q respectively, and is parallel to the
If l ‖ ̅̅̅̅ then
Corollary 76 If in , line l intersects ̅̅̅̅ and ̅̅̅̅ at P and Q respectively, and is parallel to the third side ̅̅̅̅ , then . If l
̅̅̅̅ , then
Corollary 76 , line l intersects ̅̅̅̅ and ̅̅̅̅ at P and Q, respectively and ̅̅̅̅ .
If in
A P
Q l
B
C ̅̅̅̅ .
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Theorem 76 Triangles similar to the same triangle are similar to each other.
If
.
Theorem 77 The SAS Similarity Theorem If two sides of one triangle are proportional to the corresponding two sides of another triangle and their respective included angles are congruent, then the triangles are similar. If
and âŚ&#x;B
âŚ&#x;E then
ABC
DEF
Theorem 78 The SSS Similarity Theorem If the sides of one triangle are proportional to the corresponding sides of a second triangle, then the triangles are similar. A 2
5 B
If in
D
4
4 C
10 E
8
F
then
Theorem 79 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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The L-L similarity theorem If the legs of a right triangle are proportional to the corresponding legs of another right triangle, the right triangles are similar. A
D
3
6
C
4
B
F
If in right triangles ABC and DEF,
8
E
then
Theorem 80 The H-L Similarity Theorem If the hypotenuse and a leg of one right triangle are proportional to the corresponding hypotenuse and leg of another right triangle, then the right triangles are similar. If in right triangles ABC and DEF, then A
C
D
B F
E
Postulates and theorems that can prove triangles are similar: 1. 2. 3. 4.
The AAA Similarity Postulate The AA Similarity Theorem The SAS Similarity Theorem The SSS Similarity Theorem
Right triangles can be proven similar by the: 1. L-L Similarity Theorem 2. H-L Similarity Theorem
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Practice Determine whether
. Give a reason for your answer.
1. 2. m⦟S = 40, m⦟E = 70, m⦟A = 70, and m⦟K = 80 ̅̅̅̅ ̅̅̅̅ 3. ̅̅̅̅ ̅̅̅̅̅
Lesson 5.4 Proportional Segments in Triangles
Theorem 81 The Proportional Segments Theorem If a line intersects two sides of a triangle at a distinct point and is parallel to the third side, the line divides the two sides into two proportional segments. If ̅̅̅̅
̅̅̅̅ , then
=
Theorem 82 The Converse of the Proportional Segments Theorem If a line intersects the two sides of a triangle at distinct points and cut these sides into proportional segments, then the line is parallel to the third side.
then ⃡
If in
̅̅̅̅ .
Corollary 82 A line parallel to one side of a triangle divides the other two sides and cuts off segments which are proportional to these sides. If l
̅̅̅̅ , then
and
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A D
E
l
B
C
Corollary 83 If three or more parallel lines are cut by two or more transversals, then they intercept proportional segments on these transversals. A
B
C
G
D
E
F
If ̅̅̅̅ ̅̅̅̅ ̅̅̅̅ , then Theorem 83 The bisector of an angle or a triangle divides the opposite side into segments which are proportional to adjacent sides. A
B
D
C
If ̅̅̅̅ is an angle bisector of
then
=
Theorem 84 Corresponding medians of similar triangles are proportional to the corresponding sides. and ̅̅̅̅̅
If
̅̅̅̅ are medians of
respectively, then
. Theorem 85
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Corresponding altitudes of similar triangles are proportional to the corresponding sides. S
N
G
T
I
E
̅̅̅̅
If
P
O
̅̅̅̅ are altitudes, then
Test Use the figure to complete each proportion 1. 2.
a
b
c
3. 4. 5.
x
y
z
Lesson 5.5 Some theorems on Right Triangles
Theorem 86 The altitude to the hypotenuse of a right triangle separates the right triangle into two triangles which are similar to each other and to the original triangle. If in right
̅̅̅̅ is an altitude to the hypotenuse, then
.
A D C
B
Theorem 87 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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In any right triangle: 1. The altitude to the hypotenuse is the geometric mean between the segments into which it separates the hypotenuse. 2. Each leg is a geometric mean of the hypotenuse and the hypotenuse and the segment of the hypotenuse adjacent to the leg. ̅̅̅̅ an altitude to the hypotenuse, then
If in right , A D
C
B
Theorem 88 The Pythagorean Theorem In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. If
is a right triangle, then (AB)2 = (BC)2 + (AC)2.
B
C
A
Pythagorean Theorem: if triangle ABC is a right , then c2 = a2 + b2 Pythagorean Theorem Converse: if c2 = a2 + b2, then B
B
a C
c b
a A
C
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A Pythagorean triple is a group of three whole numbers that satisfies the equation a 2 + b2 = c2 where c is the greatest number. One common Pythagorean triple is 3,4, and 5. Example Use the Pythagorean Theorem to find each missing length. All measurements are in centimeters. a. 6
x
8 Solution: a 2 + b 2 = c2 62 + 82 = x2 36 + 64 = x2 100 = x2 10 = x Theorem 90 If the square of the length of the longest side of a triangle is greater than the sum of the squares of the lengths of the other two sides, the triangle is obtuse. B a
c C
If c2 > a2 + b2, then
b
A
is obtuse.
Theorem 91 If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides, the triangle is acute. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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B a C
c b
A
If c2 < a2 + b2, then
is acute.
Other theorems of right triangles Theorem 92 The median Theorem The median of the hypotenuse of a right triangle is one half as long as the hypotenuse. If in right
, ̅̅̅̅ is the median to the hypotenuse, then CD =
.
Special Right Triangles Theorem 93 The 30-60-90 Triangle Theorem Ina 30-60-90 triangle, the side opposite the 300 angle is half as long as the hypotenuse and the side opposite the 600 angle is √ times as long as the side opposite the 300 angle. If right
with m⦟A = 30, m⦟B = 60, m⦟C = 90, then BC = AB and AC = √ BC. B h C
√
A
Theorem 94 The Converse of the 30-60-90 Triangle Theorem
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If in a right triangle one leg is half as long as the hypotenuse, then its opposite angle has a measure of 30. B 600 c 300 C If in right
BC =
A , then m⦟A = 30
Theorem 95 The isosceles Right Triangle Theorem In an isosceles right triangle, the hypotenuse is √ times as long as either of the legs. If
is an isosceles right , with AC = BC = x, AB = c, and m⦟C = 90, then c = √
.
Theorem 96 If two triangles (polygons) are similar, then the ratio of their areas is equal to the square of the ratio of – 1. Any two corresponding sides. 2. Any two corresponding altitudes. 3. Any two corresponding medians. 4. Their perimeters
Theorem 97 Areas and perimeter of Similar Figures If the similarity ratio of two similar figures is , then YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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1. The ratio of their perimeters is , and 2. The ratio of their areas is
.
Test Solve each 1. Find the length of the diagonal of a square whose sides measure 6 cm. 2. The lengths of the three sides of a right triangle are consecutive integers. Find them. 3. The diagonal of a square measures 16 meters. Find each area.
Chapter 6 Trigonometry
A trigonometric ratio is the ratio of the lengths of two sides of a right triangle. The trigonometric ratio is the ratio of the lengths of two sides of a right triangle. Sine, Cosine, and Tangent For all right triangles ABC with acute angle A Sine A =
or sin A =
Cosine A =
or cos A =
Tangent A =
or tan A =
Cosecant A = Secant A = Cotangent A =
or csc A = or sec A = or cot A =
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Hypotenuse B c
a
A
b
C
Adjacent leg to ⦟
leg opposite to ⦟A
Sines, cosines and Tangents of Special Angles Special Angles sin cos tan 300 √ √ 450
√
600
√
√
1 √
Example Find the sine, cosine, and tangent of 300 and 600. Solution Sketch a 30-60-90 triangle. To make the calculation simple, you can choose 1 as the length of the shorter leg. From the 30-60-90 triangle theorem, it follows that the length of the longer leg is √ and the length of the hypotenuse is 2. Sin 300 = Cos 300 = Tan 300 =
sin 600 = √
√
=
√
=
√
cos 600 =
=
tasn 600 =
=
Notice that sin 300 = = cos 600. Also, sin 600 =
√
√
=√
= cos 300 and sin 450 =
√
= cos 450.
This happens because 300 and 600 are complementary angles as well as 450 and 450.
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In general, it can be shown that the cofunctions of complementary angles are equal. This means that if is an acute angle, then the following relationships are true: Sin (900 – ) = cos
csc (900 – ) = cos
Cos (900 – ) = sin
sec (900 – ) = csc
Tan (900 – ) = cot
cot (900 – ) = tan
The values of trigonometric ratios of angles different from special angles can be found using a scientific calculator. For example, to find sin 70 0 press sin key then press 7 and 0 keys to get 0.93969262. Sin 70 = 0.93969262 Practice Use a calculator to find 1. 2. 3. 4.
Sin 430 Sin 500 Cos 430 Tan 850
All angles do not have measures that are an integral number of degrees, such as the 500 angle, the 430 angle, and the 850 angle. The measures of some angles are in fractions of a degree such as 65.1250. Angle measures in this form are said to be in decimal degrees. The measures of angles can also be written using degrees ( 0 ), minutes (“ ) and seconds ( “ ). Such as 65018”45”. A degree is divided into 60 minutes and a minute is divided into 60 seconds. 10 = 60’ 1’ = 60” The equations above can be used to convert an angle measure from degrees, minutes, seconds to decimal degrees. For example 65018’45” can be converted as: 65018’45” = 650 + 18( ) + 45 (
) = 65.3125
To convert the decimal degrees 65.3125 to an angle measure in degrees, minutes, seconds, we have: YOUNG JI INTERNATIONAL SCHOOL/COLLEGE
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65.3125 = 650 + 0.3125 = 650 + 18.75’ = 650 + 18 + 0.75’( ) = 650 + 18’ + 45”
Practice a. Convert 47012’20” to decimal degrees. b. Convert 24.2514 to degree, minute, second.
Using trigonometric ratios From the definitions and the given examples, the following relationships can be established: a.
, where sin , where cos , where tan
b. If we take the ratio of sin =
to cos , we have
= tan
Similarly,
=
= cot
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c. From the Pythagorean theorem, c2 = a2 + b2 or (hyp)2 = (opp)2 + (adj)2, if both sides of the equation are divided by (hyp)2 we have (
)
(
Since
)
(
)
= sin ,
, and
= 1, the following relationship holds
1 = sin2 + cos2 the following can be derived by dividing both sides of the equation (hyp.) 2 = (opp.)2 + (adj)2 by (opp)2 and (adj)2, respectively csc2 = 1 + cot2 sec2 = 1 + tan2
and
Example Express the terms of sin csc
and cos
and then simplify.
cot
Solution csc
cot
= =
Practice I.
Sketch a right triangle corresponding to the given trigonometric ratio value of . Use the Pythagorean theorem to determine the third side and then find the other five trigonometric ratio values of . 1. Cos
=
2. sec
=
3. Tan
=
4. Tan
=
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II.
Find the exact value of each expression. 1. Csc 300 – sec 300 2. Sec 600 cos 600 – tan 600 3. 2 sin 300 – sec 600Tan 450
Lesson 6.2 Trigonometric Ratios of Obtuse Angles
A reference angle is an acute angle between the terminal side and the x-axis. Angle = 300 is known as the reference angle of 1500. The angle reference angle of 2100 and 3300 angles In general, an acute angle –0
= 300 also a
is the reference angle of angles 1800 – , 180 +
and 3600
Using reference angles to evaluate trigonometric ratio 1. Find the associated reference angle, , and the trigonometric ratio value for . 2. Use the quadrant in which P( ) lies to annex the appropriate sign to the trigonometric ratio value in step 1. Example Express sin 1600 Solution Sin 1600 = sin ( 180 – 20)0 = sin 200 = 0.3420 Cos 1150 = cos ( 180 – 65)0 = -cos 650 = -0.4226
Co terminal Angles
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An angle of x0 is co terminal with angles of x0 + K
0
, where k is an integer.
Example Find a positive co terminal angle less than 3600 that is co terminal with -150 angle Solution Add 3600 to find a positive co terminal angle. -150 + 360 = 2100 A 2100 angle is co terminal with the -1500 angle.
Practice Determine two angles which are co terminal to the given angles 1. 2. 3. 4. 5.
2400 2250 -4100 500 1150
Lesson 6.3 The Law of Sines
For any triangle ABC,
Note: 1. The Law of Sines can also be written in the reciprocal form. 2. The Law of sines can also take several forms. a=
b=
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a=
b=
c=
Example Solve each triangle ABC using the given information 1. a=12 cm, b = 23 cm, and A = 340 Solution Find B using the Law of sines
Sin B = =
Practice Solve each triangle 1. If A = 420, B = 960, and b = 12 2. If A = 420, B = 480, and c = 12 3. If A = 330, C = 1280, and a = 16
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