If  $\alpha$ and  $\beta$ are the zeroes of the quadratic polynomial  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{kx}}^2+4\mathrm{x}+4$ such that ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=24$, then the values of k is

It is given that  $\alpha$ and  $\beta$ are the zeroes of the quadratic polynomial  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{kx}}^2+4\mathrm{x}+4$

$\therefore \mathrm{\alpha }+\mathrm{\beta }=-\frac{4}{\mathrm{k}}\mathrm{and},\mathrm{\alpha \beta }=\frac{4}{\mathrm{k}}$

We have, 

$\begin{array}{l}{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=24\\ \Rightarrow (\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }=24\left[\because {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }\right]\\ \Rightarrow {\left(-\frac{4}{\mathrm{k}}\right)}^2-2\times \frac{4}{\mathrm{k}}=24\\ \Rightarrow \frac{16}{{\mathrm{k}}^2}-\frac{8}{\mathrm{k}}=24\\ \Rightarrow 16-8\mathrm{k}=24{\mathrm{k}}^2\\ \Rightarrow 3{\mathrm{k}}^2+\mathrm{k}-2=0\\ \Rightarrow 3{\mathrm{k}}^2+3\mathrm{k}-2\mathrm{k}-2=0\\ \Rightarrow 3\mathrm{k}(\mathrm{k}+1)-2(\mathrm{k}+1)=0\\ \Rightarrow (\mathrm{k}+1)(3\mathrm{k}-2)=0\Rightarrow \mathrm{k}+1=0\text{ or }3\mathrm{k}-2=0\\ \Rightarrow \mathrm{k}=-1\text{ or }\mathrm{k}=\frac{2}{3}\\ \mathrm{Hence}, \mathrm{k}=-1\text{ or }\mathrm{k}=\frac{2}{3}\end{array}$