J17

x – 7 = 0 or x + 7 = 0 x = 7 or x = -7.
The roots are -7 and 7.
(b) Factorizing x 2 – 6x = 0 gives x(x – 6) = 0
Either x = 0 or x – 6 = 0
x = 0 or x = 6
The roots are 0 and 6.
(c) x 2 – 16x + 64 = 0
Factorizing x 2 – 16x + 64 = 0 gives (x – 8)(x – 8) = 0
Either x – 8 = 0 or x – 8 = 0
x = 8 or x = 8
The roots are 8 and 8.
Note: x 2 – 16x + 64 is a perfect square and therefore it has identical factors. The
equation x 2 – 16x + 64 = 0 has two equal roots.
(d) 6x 2 + 5x – 4 = 0
When the coefficient of x 2 ,(in this case it is 6), in the quadratic expression is
numerically greater than 1, we proceed as follows when factorizing:
- Multiply the coefficient of x 2 by the constant term, i.e. 6 × -4 = -24.
- Find the factors of -24 whose sum is 5, (the coefficient of x), i.e. -3 and 8.
- Rewrite the equation as:
6x 2 + (-3 + 8)x – 4 = 0
6x 2 – 3x + 8x – 4 = 0
3x(2x – 1) + 4(2x – 1) = 0
(2x – 1)(3x + 4) = 0
Then, either 2x – 1 = 0 or 3x + 4 = 0
2x = 1 or 3x = -4
x =
1
2
or x = - 4 3
.
Example
Factorize 3x 2 – 22x + 7. Hence solve 3x 2 – 22x + 7 = 0
Solution
3x 2 – 22x + 7
Multiplying 3 by 7, gives 21. The factors of 21 whose sum is -22, are -1 and
-21.
Then, 3x 2 – 22x + 7 3x 2 + [-1 + (-21)]x + 7
3x 2 – 1x – 21x + 7
x(3x – 1) -7(3x – 1)
(3x – 1)(x – 7)
Hence, 3x 2 – 22x + 7 = (3x – 1)(x – 7).
The equation 3x 2 – 22x + 7 = 0 can be written as (3x – 1)(x – 7) = 0
Either 3x – 1 = 0 or x – 7 = 0
3x = 1 or x = 7