If one root of the equation $x^{2} - 6 x + q = 0$ is twice the other, find the value of 'q'.

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Solution

Consider the equation $x^{2} - 6 x + q = 0$ , where a = 1, b = -6, c = q
(i) Sum of the roots $m + n = \frac{- b}{a} = \frac{- (- 6)}{1} ∴ m + n = 6$
(ii) Product of the roots $m n = \frac{c}{a} = \frac{q}{1} ∴ m n = q$
If one root is m then twice the root is 2m.
$∴ m = m$ and $n = 2 m$
$m + n = 6 ∴ m + 2 m = 6, 3 m = 6 ∴ m = \frac{6}{3} = 2$
We know that $q = m n ∴ q = m (2 m) = 2 m^{2} = 2 (2)^{2} = 8 ∴ q = 8$

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