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Q.
If $x^{2}+ax+10=0$ and $x^{2}+bx-10=0$ have common roots, then $a^{2} - b^{2}$ is equal to
Complex Numbers and Quadratic Equations
Solution:
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 $\ldots\left(i\right)$
$\alpha^{2}+b\alpha-10=0$ or $\alpha^{2}=10-b\alpha \, \ldots\left(ii\right)$
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 $\left(i\right)$, $\frac{400}{\left(b-a\right)^{2}}+\frac{20a}{\left(b-a\right)}+10=0$
$\Rightarrow \,10\left(b-a\right)^{2}+20a\left(b-a\right)+400=0$
$\Rightarrow \,\left(b-a\right)^{2}+2a\left(b-a\right)+40=0$
$\Rightarrow \, a^{2}+b^{2}-2ab+2ab-2a^{2}+40=0$
$\Rightarrow \, b^{2}-a^{2}+40=0$
$\Rightarrow \, a^{2}-b^{2}=40$.