SOLVE A QUADRATIC EQUATION BY FACTORING
About "Solve a quadratic equation by factoring"
Solve a quadratic equation by factoring :
Usually we have three methods to solve a quadratic equations.
(1) Factoring
(2) Using quadratic formula
(3) Completing the square method
Here we are going to see how to solve a quadratic equation by using the method factoring.
To factor a quadratic equation which is in the form ax² + bx + c = 0, we have to follow the steps given below.
Solving quadratic equations in which the value of a is 1
- If it is 1, then we have to take the constant term and split it into two factors.
- In which the product of two factors must be equal to the constant term and the simplified value of those factors equal to the middle term, that is coefficient of x.
- Write each factors in the form (x + a) (x + b) according to the sign.
- Set each factors equal to zero to get the value of x.
Solving quadratic equations in which the value of a is not 1
- Multiply the coefficient of x² and constant term.
- Split this product into two factors such that their sum is equal to the coefficient of x .
- Divide each factors by the coefficient of x².
- If it is possible we may simplify, otherwise we have to write the denominator along with x.
- write each factors in the form (x + a) (x + b).
- Set each factors equal to zero to get the value of x.
Solve a quadratic equation by factoring - Examples
Example 1 :
Solve x² + 9 x + 14 = 0
Solution :
Since the coefficient of x² is 1, split the constant term that into two parts.
14 = 2 ⋅ 7, 2 + 7 = 9
x² + 9 x + 14 = 0
(x + 2) (x + 7) = 0
x - 2 = 0 x + 7 = 0
x = 2, x = -7
Hence the solution is {2, -7}
Example 2 :
Solve x² - 9 x + 14 = 0
Solution :
Since the coefficient of x² is 1, split the constant term that into two parts.
14 = -2 ⋅ (-7) , -2 - 7 = -9
x² - 9 x + 14 = (x - 2) (x - 7)
(x - 2) (x - 7) = 0
x - 2 = 0 x - 7 = 0
x = 2, x = 7
Hence the solution is {2, 7}
Example 3 :
Solve x² + 2 x - 15 = 0
Solution :
Since the coefficient of x² is 1, split the constant term that into two parts.
-15 = -3 ⋅ 5 , -3 + 5 = 2
x² + 2 x - 15 = (x - 3) (x + 5)
(x - 3) (x + 5) = 0
x - 3 = 0 x + 5 = 0
x = 3, x = -5
Hence the solution is {3, -5}
Example 4 :
Solve x² - 2 x - 15 = 0
Solution :
Since the coefficient of x² is 1, split the constant term that into two parts.
-15 = -5 ⋅ 3 , -5 + 3 = -2
x² - 2 x - 15 = (x - 5) (x + 3)
(x - 5) (x + 3) = 0
x - 5 = 0 x + 3 = 0
x = 5, x = -3
Hence the solution is {-3, 5}
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