1. Evaluate each expression without using a calculator. (a)
−34 =−32 −32 =9 9 =81 (b)
(c)
(d)
−34 =−13 2 32 =−19 9 =−81 1 34 1 = 2 2 3 3 1 = 9⋅9 1 = 81
3−4 =
523 23 −21 21 =5 5 5 =5
23−21
=52 =25
(e )
2 −2 2−2 = −2 3 3 32 = 2 2 3⋅3 = 2⋅2 9 = 4
(f ) −3 / 4
16
4
= 16 1 =4 3 16
−3
1 16⋅16⋅16 1 4 1 4 1 =4 16 16 16 4 4 4 1 1 1 = 4 4 4 16 16 16 1 1 1 = 2 2 2 1 = 8 =
4
2. Simplify each expression. Write your answer without negative exponents. (a)
(b)
200− 32= 100⋅2− 8⋅4 = 25⋅4⋅2− 4⋅2⋅4 = 25 4 2− 4 2 4 =52 2−2 2 2 =10 2−4 2 =6 2 3a 3 b3 4ab 2 2 =3a 3 b3 4ab 2 4ab 2 =3a 3 b3 4 2 a2 b2 2 =3a 3 b3 16⋅a2⋅b2⋅2 =3a 3 b3 16⋅a2⋅b4 =3⋅16a 3⋅a 2 b3⋅b 4 32 34 =48 a b =48⋅a5⋅b7 5 7 =48a b
(c)
3/ 2 3 −2 3x 3/ 2 y 3 −2 3x y 2 −1 / 2 = 2 −1/ 2 −2 x y x y 2
−1/ 2 2
x y = 3x 3/ 2 y 3 2 x 2⋅2 y
=
2
3 x
3 ⋅2 2
−1 ⋅2 2
y 3⋅2
x 4 y−1 = 3 6 9x y x 4 x−3 = 9 y 6 y 1 x 4−3 = 9 y 61 1
x = 7 9y =
x 9y 7
3. Expand and simplify. (a)
3 x64 2x−5=3x188x−20 =11x−2 (b)
x3 4x−5= x⋅4x− x⋅53⋅4x−3⋅5 =4x 2−5x12x−15 2 =4x 7x−15
(c)
a b a− b= a⋅ a− a⋅ b b⋅ a− b⋅ b 1 2
1 2
1 2
1 2
1 2
1 2
1 2
= a ⋅ a −a ⋅b b ⋅ a −b ⋅b =a
1 1 2 2
1 2
1 2
=a
0−b
2 2
=a −b
2 2
=a1−b1 =a−b
1 2
− a ⋅ b a ⋅b −b =0
11 2
1 2
11 2
1 1 2 2
1 2
(d)
2x32 = 2x32x3 =2x⋅2x2x⋅33⋅2x3⋅3 =4x 26x6x9 =4x 212x9 (e)
x23= x22 x2 = x2 x2 x2 = x⋅x x⋅22⋅x2⋅2 x2 = x 2 2x2x4 x2 = x 24x4 x2 2 2 = x ⋅x x ⋅24x⋅x4x⋅24⋅x4⋅2 = x 3 2x 2 4x 28x4x8 = x 3 6x 212x8
4. Factor each expression. ( a ) Factoring using the pattern for the differences of squares: a2 - b2 = (a - b)(a + b)
2
4x −25=? 4x 2 −25= 2x−52x5
(b)
2x 25x−12= 2x−3 x4 (c)
x 3 −3x 2−4x12= x 3 −3x 2 −4x12 = x 2 x−3−4 x−3 = x 2−4 x−3 = x2 x−2 x−3 (d)
x 427x= x x 3 27 = x x 3 33 = x x3 x 2 −3x32 = x x3 x 2 −3x9
(e)
3 2
1 2
−
1 2
3 2
1 2
3x −9x 6x =3x −9x
6 x
=
x x
=
1 2
3 2
=x =x
1 2
−
1 2
3x −9x 6x
1 2
3 2
1 x
1 2
1 2
1 2
1 2
3x x −9x x 6x
1 2
−
1 2
−
1 2
−
1 2
3x
3 1 2 2
−9x
1 1 2 2
3x 2 −9x6 2
= x ⋅3 x −3x2 =3 x
−
1 2
x−1 x−2
(f)
x 3 y−4xy= yx x 2 −4 = yx x2 x−2
6x
−
1 2
1 2
x
−1 1 2 2
5. Simplify the rational expression. (a) 2 x 3x2 x1 x2 = 2 x − x−2 x−2 x1 x2 = x−2
(b)
2x 2 − x−1 x3 2x1 x−1 x3 ⋅ = ⋅ 2 x −9 2x1 x−3 x3 2x1 =
x−1 x−3
(c)
x2 x1 x2 x1 − = − 2 x −4 x2 x−2 x2 x2 x 2− x−2 x1 = x−2 x2 x 2− x 2 x−2x−2 = x−2 x2 2
2
x − x − x2x2 = x−2 x2 =
x2 x−2 x2
=
1 x−2
(d)
y x y 2− x2 − x y xy = 1 1 x− y − y x xy 2
2
xy y − x = xy x− y
=
y− x y x x− y
=
y− x y x − y− x
=− yx
6. Rationalize the expression and simplify. (a)
10 = 10 ⋅ 52 5−2 5−2 52 =
10 5 10 2 5 5 5⋅2−2 5−4
=
10⋅52 10 2 5 2 5−2 5−4
=
2⋅5⋅52 10 1 2 2
5 −4 =
2⋅252 10
=
5−4 2 252 10 1
=5 22 10
(b)
4h−2 = 4h−2⋅ 4h2 h h 4h2 =
4h−2 4h2 h 4h2 2
4h −2⋅2 = h 4hh⋅2 1 2 2
=
4h −4 h 4h2h
=
4h−4 h 4h2h
=
h h 4h2h
=
h h 4h2
=
1 4h2
7. Rewrite by completing the square. (a) This is the original problem.
2
x x1=0 2
Move the loose number over to the other side.
x x=−1
Take half of the coefficient of x (don't forget the sign!) of the x-term
x=1 x 1 = 2 2
...and square it .
x 1 = 2 4
2
Add this square to both sides of the equation.
1 1 x 2 x =−1 4 4
Convert the left-hand side to squared form, and simplify the right-hand side. (This is where you use that sign that you kept track of earlier. You plug it into the middle of the parenthetical part.)
1 3 x =− 2 4
Move the loose number over to the other original side.
1 2 3 x =0 2 4
(b) This is the original problem.
2
2
2x −12x11=0 2
Move the loose number over to the other side.
2x −12x=−11
Factorize the common number.
2 x 2−6x=−11 −11 x 2−6x= 2
Divide through by whatever is multiplied on the squared term.
Take half of the coefficient of x (don't forget the sign!) of the x-term
−6 −6 =3 2
...and square it . Add this square to both sides of the equation.
Convert the left-hand side to squared form, and simplify the right-hand side. (This is where you use that sign that you kept track of earlier. You plug it into the middle of the parenthetical part.) Move the loose number over to the other original side. Multiply for the same number of the denominator that is in the loose number all the equation. And the result is:
2
3 =9 x 2−6x9= x−32=
−11 9 2
7 2
7 x−32− =0 2 7 2 x−32 − =0 2 2
2 x−3 −7
8. Solve the equation. (Find only the real solutions.) (a)
x5=14−
1 2
1 x5−5=14− −5 2 1 x=14− −5 2 x=
17 2
(b)
2x 2x−1 = x1 x 2x x= 2x−1 x1 2
2
2x =2x 2x− x−1 2x 2 −2x 2 −2x x=−1 −2x x=−1 − x=−1 −1 x=−1 x=
−1 −1
x=1 (c) 2
x − x−12=0 x−4 x3=0 Roots x=4 x=−3
(d)
2
2x 4x1=0 1 2 2 x 2x =0 2 1 x 22x =0 2 1 x 22x=− 2 1 2 x 2x1=− 1 2 x 22x1= x12 =
−12 2
1 2
Square-root both sides, remembering the "±" on the right-hand side. Simplify as necessary.
x12=
x1=±
1 2
1 2
Solve for "x =".
x=−1± Remember that the "±" means that you have two values for x.
1 2
1 2 1 x 2=−1− 2 x 1 =−1
Then we are going to rationalize the expression and simplify.
1 1 = 2 2 =
=
1 2 2 2
=
=
=
Then substitute the result above in the roots.
1 1 2
1 2 2 2
1⋅ 2 2
2
2 2
1 = ⋅ 2 2 1 x 1 =−1 ⋅ 2 2 1 x 2=−1− ⋅ 2 2
(e) 4
2
This polynomial is not quadratic, it has degree four. However, it can be thought of as quadratic in x2.
x −3x 2=0
However, it can be thought of as quadratic in x2.
x −3 x 2=0
It might help you to actually substitute z for x2.
z −3z2=0
This is a quadratic equation in z.
2 2
2
2
z−1z −2=0 z 1=1 We are not done, because we need to find values of x that make the original equation true. Now replace z by x2 and solve the resulting equations
z 2 =2
2
x =1 x 1 =1
x 2 =−1
2
x =2 x 3= 2 So there are four solutions, four real.
x 4 =− 2
(f)
First, I'll isolate the absolutevalue part; that is, I'll get the absolute-value expression by itself on one side of the "equals" sign, with everything else on the other side:
Now I'll clear the absolutevalue bars by splitting the equation into its two cases, one for each sign:
3∣x−4∣=10 10 ∣x−4∣= 3 x−4 =
x−4=
10 3
10 4 3 1012 x= 3 22 x= 3 x=
So the solution is
10 3
x 1=
22 3
10 3 10 − x4= 3 10 − x= −4 3 10−12 − x= 3 2 − x=− 3 2 − 3 x= −1 −2 x= −13 2 x= 3 2 x 2= 3 − x−4=
(g)
1 − 2
2x 4− x
−3 4− x=0
1 − 2
2x 4− x
=3 4− x
4− x
2x=3
1 − 2
4− x
1 2
2x=3 4− x 4− x 2x=3 4− x 2x=12−3x 2x3x=12 5x=12 x=
12 5
9. Solve each inequality. Write your answer using interval notation. (a)
−45−3x17 −4−55−3x−517−5 −4−5−3x17−5 −9−3x12 −
9 12 x −3 −3
3 x−4 Interval:
[-4, 3)
(b)
x 22x8 2
x −2x8 x 2−2x181 x−12 9
x−12± 9 −3 x−13 −31 x31 −2 x4 (c)
x x−1 x20 x x 2 2x−1x−20 x x 2 x−20 x 3 x 2 −2x0
x
x (x - 1)(x + 2)
Is it greater than zero? ¿Es mayor que cero?
Intervals "Intervalo"
2
2 (2 – 1) (2 + 2) = 2 * 1 *4=8
Yes
(1, infinity)
0.5
½ (½ -1)(1/2 + 2) = ½ (- No ½ )(5/2 ) = - 5/ 8
(0, 1)
-1
-1 (-1 -1)(-1 + 2) = -1(- Yes 2)(1) = 2
(-2, 0)
-3
-3 (-3 – 1)(-3 + 2) = -3 * No (-4)(-1)= -12
(infinity, -2)
Interval : (-2, 0) U (1, infinity )
(d)
∣x−4∣3 x−43 x34 x7 − x−43 − x43 − x3−4 − x−1 x1
Interval: (1, 7)
(e)
2x−3 1 x1
2x−3 −10 x1
2x−3− x1 0 x1 2x− x−3−1 0 x1 x−4 0 x1
x= 4 x= -1
Origina intervals (- ∞ , - 1] [-1 , 4 ] [ 4, ∞ ) Evaluating the next x values to :-2, 0, 5 Sing of y
+
-
+
X-4
-
-
+
X+1
-
+
+
-1 Interval: (- 1 , 4]
4
10. State whether each equation is true or false. (a)
pq2 = p 2q2 = FALSE
(b)
ab= a b=TRUE
(c)
a2b2=ab= FALSE
(d)
1TC =1T = FALSE C
(e)
1 1 1 = − = FALSE x− y x y
(f)
1 x a b − x x
=
1 =TRUE a−b
Theory: Solving Inequalities: An Overview http://www.purplemath.com/modules/ineqsolv3.htm Rational Exponent Rules http://www.brightstorm.com/math/algebra-2/roots-and-radicals/rules-for-rational-exponents Cool math's Algebra Practice Problems - Polynomials: Simplifying with exponent rules, adding, subtracting, multiplying, dividing, factoring trinomials and more http://www.coolmath.com/algebra/algebra-practice-polynomials.html Fractional Exponents http://www.tpub.com/math1/8c.htm Inequalities http://www.sosmath.com/algebra/inequalities/ineq01/ineq01.html Solving Absolute-Value Equations http://www.purplemath.com/modules/solveabs.htm Completing the Square: Solving Quadratic Equations http://www.purplemath.com/modules/sqrquad.htm Rationalizing Denominators http://www.regentsprep.org/Regents/math/algtrig/ATO3/rdlesson.htm Pauls Online Notes : Algebra/Trig Review – Rationalizing http://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/Rationalizing.aspx