The value of $k$ for which zeros of polynomial $k x^{2} + 4 x + 4$ are $a$ and $b$ are related as $a^{2} + b^{2} = 24$ is....

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Solution

$a, b$ are the zeros of $f (x) = k x^{2} + 4 x + 4 = 0$ and $a^{2} + b^{2} = 24$
$a + b = \frac{- 4}{k}$ and $a \times b = \frac{4}{k}$

Now,
$a^{2} + b^{2} = 24$
$(a + b)^{2} - 2 a b = 24$
$(\frac{- 4}{k})^{2} - 2 \times \frac{4}{k} = 24$
$\frac{16}{k^{2}} - \frac{8}{k} = 24$
$\frac{(16 - 8 k)}{k^{2}} = 24$
$16 - 8 k = 24 k^{2}$
$24 k^{2} + 8 k - 16 = 0$
$8 (3 k^{2} + k - 2) = 0$
$3 k^{2} + k - 2 = 0$
$3 k^{2} + 3 k - 2 k - 2 = 0$
$3 k (k + 1) - 2 (k + 1) = 0$
$(k + 1) (3 k - 2) = 0$

Hence, value of $k$ is either $- 1$ or $\frac{2}{3} .$

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