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Factorizing x 2 – 16x + 64 = 0 gives (x – 8)(x – 8) = 0<br />
Either x – 8 = 0 or x – 8 = 0<br />
x = 8 or x = 8<br />
The roots are 8 and 8.<br />
Note: x 2 – 16x + 64 is a perfect square and therefore it has identical factors. The<br />
equation x 2 – 16x + 64 = 0 has two equal roots.<br />
(d) 6x 2 + 5x – 4 = 0<br />
When the coefficient of x 2 ,(in this case it is 6), in the quadratic expression is<br />
numerically greater than 1, we proceed as follows when factorizing:<br />
- Multiply the coefficient of x 2 by the constant term, i.e. 6 × -4 = -24.<br />
- Find the factors of -24 whose sum is 5, (the coefficient of x), i.e. -3 and 8.<br />
- Rewrite the equation as:<br />
6x 2 + (-3 + 8)x – 4 = 0<br />
6x 2 – 3x + 8x – 4 = 0<br />
3x(2x – 1) + 4(2x – 1) = 0<br />
(2x – 1)(3x + 4) = 0<br />
Then, either 2x – 1 = 0 or 3x + 4 = 0<br />
2x = 1 or 3x = -4<br />
x =<br />
1<br />
2<br />
or x = - 4 3<br />
.<br />
Example<br />
Factorize 3x 2 – 22x + 7. Hence solve 3x 2 – 22x + 7 = 0<br />
Solution<br />
3x 2 – 22x + 7<br />
Multiplying 3 by 7, gives 21. The factors of 21 whose sum is -22, are -1 and<br />
-21.<br />
Then, 3x 2 – 22x + 7 3x 2 + [-1 + (-21)]x + 7<br />
3x 2 – 1x – 21x + 7<br />
x(3x – 1) -7(3x – 1)<br />
(3x – 1)(x – 7)<br />
Hence, 3x 2 – 22x + 7 = (3x – 1)(x – 7).<br />
The equation 3x 2 – 22x + 7 = 0 can be written as (3x – 1)(x – 7) = 0<br />
Either 3x – 1 = 0 or x – 7 = 0<br />
3x = 1 or x = 7<br />
x = 3<br />
1<br />
or x = 7<br />
Exercise : Solve.<br />
1. (x + 4)(x + 2) = 0 2. (x – 5)(x – 3) = 0<br />
3. (x – 5)(x + 7) = 0 4. (x – 9) 2 = 0