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If $x^2+a x+10=0$ and $x^2+b x-10=0$ have a common root, then $a^2-b^2$ is equal to
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The correct answer is:
$40$
Let $\alpha$ be a common root, then
$\alpha^2+a \alpha+10=0$ ...(i)
and $\alpha^2+b \alpha-10=0$ ...(ii)
form (i) - (ii),
$(a-b) \alpha+20=0 \Rightarrow \alpha=-\frac{20}{a-b}$
Substituting the value of $\alpha$ in (i), we get
$\left(-\frac{20}{a-b}\right)^2+a\left(-\frac{20}{a-b}\right)+10=0$
$\Rightarrow 400-20 a(a-b)+10(a-b)^2=0$
$\Rightarrow 40-2 a^2+2 a b+a^2+b^2-2 a b=0$
$\Rightarrow a^2-b^2=40$.
$\alpha^2+a \alpha+10=0$ ...(i)
and $\alpha^2+b \alpha-10=0$ ...(ii)
form (i) - (ii),
$(a-b) \alpha+20=0 \Rightarrow \alpha=-\frac{20}{a-b}$
Substituting the value of $\alpha$ in (i), we get
$\left(-\frac{20}{a-b}\right)^2+a\left(-\frac{20}{a-b}\right)+10=0$
$\Rightarrow 400-20 a(a-b)+10(a-b)^2=0$
$\Rightarrow 40-2 a^2+2 a b+a^2+b^2-2 a b=0$
$\Rightarrow a^2-b^2=40$.
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