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If the roots of $a x^{2}+b x+c=0$ are $\sin \alpha$ and $\cos \alpha$ for some $\alpha$, then which one of the following is correct?
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The correct answer is:
$b^{2}-a^{2}=2 a c$
Let $\sin \alpha$ and $\cos \alpha$ be the roots of $a x^{2}+b x+c=0$ Now, $\sin \alpha+\cos \alpha=\frac{-b}{a}$ and $\sin \alpha \cos \alpha=\frac{c}{a}$
Consider $\sin \alpha+\cos \alpha=\frac{-b}{a}$
Squaring both side, $(\sin \alpha+\cos \alpha)^{2}=\frac{b^{2}}{a^{2}}$
$\Rightarrow \sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}$
$\Rightarrow 1+\frac{2 c}{a}=\frac{b^{2}}{a^{2}}$
$\Rightarrow \frac{a+2 c}{a}=\frac{b^{2}}{a^{2}} \Rightarrow a+2 c=\frac{\mathrm{b}^{2}}{\mathrm{a}}$
$\Rightarrow a^{2}+2 a c=b^{2}
\Rightarrow b^{2}-a^{2}=2 a c$
Consider $\sin \alpha+\cos \alpha=\frac{-b}{a}$
Squaring both side, $(\sin \alpha+\cos \alpha)^{2}=\frac{b^{2}}{a^{2}}$
$\Rightarrow \sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}$
$\Rightarrow 1+\frac{2 c}{a}=\frac{b^{2}}{a^{2}}$
$\Rightarrow \frac{a+2 c}{a}=\frac{b^{2}}{a^{2}} \Rightarrow a+2 c=\frac{\mathrm{b}^{2}}{\mathrm{a}}$
$\Rightarrow a^{2}+2 a c=b^{2}
\Rightarrow b^{2}-a^{2}=2 a c$
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