Question

If $\alpha$ & $\beta$ are the zeroes of $f\left(x\right)=k{x}^2-14x+4$ such that ${\alpha }^2+{\beta }^2=24$. Find $k$.

Answer 1

$f\left(x\right)=k{x}^2-14x+4$

$\Rightarrow \alpha ,\beta$ roots $\&$ ${\alpha }^2+{\beta }^2=24$

$\Rightarrow \alpha +\beta =\frac{14}{k},$ $\alpha \beta =\frac{4}{k}$

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

$\Rightarrow 24=\frac{{14}^2}{k^2}-2\left(\frac{4}{k}\right)=\frac{196}{k^2}-\frac{8}{k}=\frac{196-8k}{k^2}$

$\Rightarrow 24k^2+8k-196=0$

Dividing the function in ${}^{\prime }{k}^{\prime }$ by $4$

$\Rightarrow 6k^2+2k-49=0$

solving for $k$ by,

$\Rightarrow k=\frac{-2\pm \sqrt{{\left(82\right)}^2-4\left(6\right)(-49)}}{2\left(6\right)}$

$=\frac{-2\pm \sqrt{4+4\left(6\right)\left(49\right)}}{12}$

$=\frac{-2\pm \sqrt{(1+6\times 49)}}{12}$

$\Rightarrow k=\frac{-2\pm 2\sqrt{295}}{12}$

Hence, the value of K is $\frac{-2\pm 2\sqrt{295}}{12}.$