Question 1(iv)
Find the roots of the following quadratic equations by factorisation:
(iv) $2 x^{2} - x + \frac{1}{8} = 0$

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Solution

(iv)
$2 x^{2} - x + \frac{1}{8} = \frac{1}{8} (16 x^{2} - 8 x + 1)$
(Splitting the middle term -8x as -4x-4x because $- 4 x \times - 4 x = 16 x^{2}$ )

$= \frac{1}{8} (16 x^{2} - 4 x - 4 x + 1) = \frac{1}{8} (4 x (4 x - 1) - 1 (4 x - 1)) = \frac{1}{8} (4 x - 1)^{2}$
Roots of this equation are the values for which $(4 x - 1)^{2} = 0$
$∴ (4 x - 1) = 0$
$\Rightarrow x = \frac{1}{4}$
So both the roots of the equation is $\frac{1}{4}$

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