Mathematics - Quadratic Equation Question with Solution | TestHub

Given $a{x}^2-2bx+15=0 ...\left(\mathrm{i}\right)$ 

Has repeated roots So $D=0$

$4b^2-4\times 15\times a=0$ 

$\Rightarrow$ $b^2=15a ...\left(\mathrm{ii}\right)$

Also given 

Now $\alpha$ will satisfy both quadratic

$a{x}^2-2bx+15=0 \& x^2-2bx+21=0$

Putting the value we get

$$\begin{gathered} a{\alpha }^2-2b\alpha +15=0 \\ {\alpha }^2-2b\alpha +21=0 \\ - + - \\ ---------- \\ \left(a-1\right){\alpha }^2=6 \end{gathered}$$

${\alpha }^2=\frac{6}{a-1}$

Now in equation  (1) product of Root ${\alpha }^2=\frac{15}{a}$

So $\frac{15}{a}=\frac{6}{a-1}$$\Rightarrow 2a=5a-5\Rightarrow a=\frac{5}{3}$

Now $b^2=15a\Rightarrow b^2=15\times \frac{5}{3}$ $\Rightarrow b^2=25$

So $b=\pm 5$

Now in quadratic $x^2-2bx+21=0$ 

Putting the value of $b$ we get

$x^2-10x+21=0$  $\Rightarrow \left(x-7\right)\left(x-3\right)=0$

So $x=3$ or $7$.

Or 

$x^2+10x+21=0\Rightarrow x=-3 or x=-7$

So  $\alpha =\pm 3 \& \beta =\pm 7$

So ${\alpha }^2+{\beta }^2=3^2+7^2=9+49=58$