If one zero of the polynomial $p (x) = (k + 4) x^{2} + 13 x + 3 k$ is reciprocal of the other, then k is equal to

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Solution

The correct option is A 2
Let one zero be $t$ other be $\frac{1}{t}$

Product = $t \times \frac{1}{t} = 1$

we know that for $a x^{2} + b x + c = 0$ the product of roots $= \frac{c}{a}$

Here $t, \frac{1}{t}$ are the roots of $p (x) = (k + 4) x^{2} + 13 x + 3 k = 0$

$a = k + 4, b = 13, c = 3 k$

Product of roots = $\frac{c}{a} = \frac{3 k}{k + 4} = 1$

$3 k = k + 4$

$2 k = 4$

$∴ k = 2$

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