J17

Example
Factorize x 2 + 7x + 6 = 0. Hence solve the equation.
Solution
x 2 + 7x + 6 = 0
Factorizing, x 2 +7x + 6, gives (x + 6)(x + 1) = 0.
Therefore, x + 6 = 0 or x + 1 = 0; which means x = -6 or x = -1
These are the only two values of x which satisfy the equation x 2 + 7x + 6 = 0. We can
check these solutions by substituting each of them in the equation.
Thus, when x = -6,
(-6) 2 + 7(-6) + 6 = 36 – 42 + 6 = 0
And when x = -1
(-1) 2 + 7(-1) + 6 = 1 – 7 + 6 = 0
Note: Every quadratic equation has two solutions.
Example
Solve: x 2 + x – 72 = 0
Solution
x 2 + x – 72 = 0
(x – 8)(x + 9) = 0. x – 8 = 0 or x + 9 = 0
x = 8 or x = -9.
Example
Solve: x 2 – x – 29 = 1
Solution
Always ensure that the quadratic expression is equated to zero. This is the only time the
method used in the examples above can apply.
Thus, x 2 – x – 29 = 1 should be rewritten as x 2 – x – 29 – 1 = 0. That is,
x 2 – x – 30 = 0.
The factors of 30, whose sum is 1, are -6 and 5.
Therefore, (x – 6)(x + 5) = 0.
Either x – 6 = 0 or x + 5 = 0
x = 6 or x = -5
The roots are -5 and 6.
Example
Solve:
(a) x 2 – 49 = 0 (b) x 2 – 6x = 0
(c) x 2 – 16x + 64 = 0 (d) 6x 2 + 5x – 4 = 0
Solutions
(a) x 2 – 49 = 0 can be written as x 2 – 7 2 = 0
(x – 7)(x + 7) = 0