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Q.

If one root of $x^{2} + K x + 27 = 0$ may be triple the other, then K =

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a

$\pm 10$

b

$\pm 6$

c

$\pm 8$

d

$\pm 12$

answer is C.

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Detailed Solution

Given m:n = 1:3 where m,n are roots of the equation.

We have $\frac{{(m + n)}^{2}}{m n} = \frac{b^{2}}{a c}$

$\Rightarrow \frac{16}{3} = \frac{K^{2}}{27}$

$\Rightarrow K^{2} = 144$

$\Rightarrow K = \pm 12$

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