Let a,b∈ R be such that the equation ax^2-2bx+15=0 has repeated root α and if α and β are the roots of the equation x^2-2bx+21=0, then α ^2+β ^2 is…

Given $a{x}^{2}-2bx+15=0...(i)$ax2−2bx+15=0...(i)

Has repeated roots So $D=0$D=0

$$4{b}^{2}-4\times 15\times a=0$$4b2−4×15×a=0

$\Rightarrow$⇒ ${b}^{2}=15a...(\mathrm{ii})$b2=15a...(ii)

Also given

Now $\alpha$α will satisfy both quadratic

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

Putting the value we get

$$a{\alpha }^{2}-2b\alpha +15=0 {\alpha }^{2}-2b\alpha +21=0 -+- ---------- (a-1){\alpha }^{2}=6$$aα2−2bα+15=0α2−2bα+21=0−+−−−−−−−−−−−(a−1)α2=6

$${\alpha }^{2}=\frac{6}{a-1}$$α2=a−16​

Now in equation (1) product of Root ${\alpha }^{2}=\frac{15}{a}$α2=a15​

So $\frac{15}{a}=\frac{6}{a-1}$a15​=a−16​$\Rightarrow 2a=5a-5\Rightarrow a=\frac{5}{3}$⇒2a=5a−5⇒a=35​

Now ${b}^{2}=15a\Rightarrow {b}^{2}=15\times \frac{5}{3}$b2=15a⇒b2=15×35​ $\Rightarrow {b}^{2}=25$⇒b2=25

So $b=\pm 5$b=±5

Now in quadratic ${x}^{2}-2bx+21=0$x2−2bx+21=0

Putting the value of $b$b we get

${x}^{2}-10x+21=0$x2−10x+21=0 $\Rightarrow (x-7)(x-3)=0$⇒(x−7)(x−3)=0

So $x=3$x=3 or $7$7.

Or

$${x}^{2}+10x+21=0\Rightarrow x=-3orx=-7$$x2+10x+21=0⇒x=−3orx=−7

So \alpha =\pm 3&\beta =\pm 7

So ${\alpha }^{2}+{\beta }^{2}={3}^{2}+{7}^{2}=9+49=58$α2+β2=32+72=9+49=58