If one roots of the equation $x^{2} - 12 x + 3 k = 0$ is square of the other , then k is

3

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9

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6

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12

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Solution

The correct option is B $9$
Let one root be a and other be $a^{2}$
$a + a^{2} = 12$ and $a^{3} = 3 k$
$a^{2} + a - 12 = 0$
$(a + 4) (a - 3) = 0$
$∴ a = - 4, 3$
putting $a = 3$ in $a^{3} = 3 k$
$k = 9$

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