Question

If $\alpha$ and $\beta$ are zeros of polynomial $k{x}^2+4x+4$ and ${\alpha }^2+{\beta }^2=24$, find $k$.

Answer 1

$k{x}^2+4x+4=0$

$\alpha +\beta =\frac{-4}{k}...\left(1\right)$ $\alpha \beta =\frac{4}{k}...\left(ii\right)$

${\alpha }^2+{\beta }^2=24$

$(\alpha +\beta {)}^2-2\alpha \beta =24$

$\frac{4^2}{k^2}-2\times \frac{4}{k}=24$

$16-8k=24k^2$

$24K^2+8k-16=0$

$3k^2+k-2=0$

$3k^2+3k-2k-2=0$

(3k-2(k+1)=0)

k=-1 OR k=$\frac{2}{3}$