Solve Quadratic Equation Learn about solving quadratic equations by factoring methods. A quadratic equation is a polynomial equation of degree 2. It can be simply represented in the form: ax2 + bx + c = 0 where 'x' is the constant, a, b, c are the constant co efficient (which are not equal to 0). Method for Solving Quadratic Equations by Factoring Here is the method to solving quadratic equations by factoring In this method, we use the basic idea of factoring the co efficient 'b' in the manner shown: Know More About Multiply Fractions Calculator
b=a×c 'b' should be broken in such a way that its original value turns out to b equal to the product of 'a' and 'c'. Let us take a general description of the method by solving the generalized equation. Assumption: Let 'p' and 'q' be the constants such that b=p+q Also we know the basic requirement for the application of this method, which is b=a×c Now to proceed further, we break 'b' such that it follows the above mentioned two conditions. Taking the general equation now : ax2 + bx + c = 0 ax2 + px + qx +c = 0 (mentioned above: b = p+q) Now since we have also followed the condition b = a × c Learn More About Area Of Circle Calculator
So 'a' tuns out to be a factor of 'p' and 'q' turns out to be a factor of 'c' also. Moving forward and taking the common terms out we get : x(x + p) + q(x + p) = 0 (x + p) Ă— (x + q) = 0 Where 'p' and 'q' are the factors of the equation. To make the understanding further clear lets take an example How to Solve Quadratic Equations by Factoring Below is the example how to solve quadratic equations by factoring method Let us consider the following quadratic equation : x2 + x + 6 = 0 here a = 1, b = 1, c = 6. Now we break 'b = 1' such that the who parts turn out to be the product of 'a Ă— c' i.e. equal to 6. x2 + 3x - 2x + 6 = 0 ( both of our necessary conditions are fulfilled here)
Now taking out the common terms gives x(x + 3) - 2(x + 3) = 0 This finally yields (x + 3)(x - 2) = 0 Here x = -3, 2 are the factors of the given equation.
Examples of Independent Variables n this page we are going to discuss about independent variable. Below you can see how we define independent variable. Independent variable definition In an algebraic equation, independent variable means a variable whose values are Independent of changes. In the values of other variables. If y is dependent variable then x is said to be independent variables. For example, take x and y are two variable in the given algebraic equation. Here every value of x is definitely connected with any other value of y, then y value is said to be function of x value is called as an independent variable and y value is called as a “dependent� variable. y values are depending upon the values of x values. Therefore, y = x2 It means y is a dependent variable, Is the square of x value is independent variable.
What is an Independent variable ? In a word, Independent variable means a variable whose it determine the values of other variables. Examples of Independent Variable Below are the examples on Independent Variable Example 1 : Solve the equation 5x+4y +8 =0 Fine the coordinate of vertexes of lines and the x axis in the polynomial. Solution: Given 5x+4y +8 =0 Substitute x = 0 in the given equation we get, 5(0) +4y +8=0 4y+8=0
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Subtract 8 on both sides we get , 4y + 8 - 8 = 0 - 8 4y = -8 Divide 4 on both sides we get y value, y = -2 Therefore the point is (0, -2) Now,plug-in x = 1 in the given equation, Exanple 2 : Solve the equation 3x-3y +12 =0 Fine the coordinate of vertexes of lines and the x axis in the polynomial. Solution: Given:3x-3y +12 =0 Plug-in x=0 in the given equation; 3(0) -3y = -12
-3y = -12 Y= 4 Plug-in x = 1 in the given equation; 3(1) - 3y = -12 -3y = -12 -3 -3y = -15 Y=5 Plug-in x=2 in the given equation; 3(2) - 3y = -12 6 - 3y = -12 -3y = -12 -6 -3y = -18 Y=6
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