Question

If $x^2+ax+10=0$ and $x^2+bx-10=0$ have common roots, then $a^2-b^2$ is equal to

• A. $10$
• B. $20$
• C. $30$
• D. $40$

Answer 1

Answer: (D)

We have, $x^2+ax+10=0,x^2+bx-10=0$
Let $\alpha$ be a common root of both the equations
So ${\alpha }^2+a\alpha +10=0$ and $\dots \left(i\right)$
${\alpha }^2+b\alpha -10=0$ or ${\alpha }^2=10-b\alpha \dots \left(ii\right)$
Putting $\left(ii\right)$ in $\left(i\right)$, we get $\left(10-b\alpha \right)+a\alpha +10=0$
$\Rightarrow \alpha \left(a-b\right)=-20$
or $\alpha =\frac{20}{b-a}$
Substituting in $\left(i\right)$, $\frac{400}{{\left(b-a\right)}^2}+\frac{20a}{\left(b-a\right)}+10=0$
$\Rightarrow 10{\left(b-a\right)}^2+20a\left(b-a\right)+400=0$
$\Rightarrow {\left(b-a\right)}^2+2a\left(b-a\right)+40=0$
$\Rightarrow a^2+b^2-2ab+2ab-2a^2+40=0$
$\Rightarrow b^2-a^2+40=0$
$\Rightarrow a^2-b^2=40$.