if $α a n d β$ are the zeroes of the quadratic polynomial $f (x) = {kx}^{2} + 4 x + 4$ such that $α^{2} + β^{2} = 24$ then the values of k is

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$\frac{2}{3}$

-1

Both 1 and 3

$\frac{3}{2}$

answer is D.

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Detailed Solution

It is given that $α$ and $β$ are the zeroes of the quadratic polynomial $f (x) = {kx}^{2} + 4 x + 4$
$∴ α + β = - \frac{4}{k} and, αβ = \frac{4}{k}$
We have,
$α^{2} + β^{2} = 24$
$\begin{array}{ll} \Rightarrow (α + β)^{2} - 2 αβ = 24 [∵ α^{2} + β^{2} = (α + β)^{2} - 2 αβ] \\ \Rightarrow {(- \frac{4}{k})}^{2} - 2 \times \frac{4}{k} = 24 \Rightarrow \frac{16}{k^{2}} - \frac{8}{k} = 24 \\ \Rightarrow 16 - 8 k = 24 k^{2} \\ \Rightarrow 3 k^{2} + k - 2 = 0 \\ \Rightarrow 3 k^{2} + 3 k - 2 k - 2 = 0 \\ \Rightarrow 3 k (k + 1) - 2 (k + 1) = 0 \\ \Rightarrow (k + 1) (3 k - 2) = 0 \Rightarrow k + 1 = 0 or 3 k - 2 = 0 \\ \Rightarrow k = - 1 or k = \frac{2}{3} \end{array}$
Hence, $k = - 1 or k = \frac{2}{3}$