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If the roots of the equation $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+\left(b^{2}+c^{2}\right)=0$ are equal, then
which one of the following is correct?
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which one of the following is correct?
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Verified Answer
The correct answer is:
$b^{2}=a c$
Since, the roots of the equation $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+\left(b^{2}+c^{2}\right)=0$ are equal.
therefore, discriminant $=0$
$\Rightarrow[2 b(a+c)]^{2}-4\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)=0$
$4 b^{2}(a+c)^{2}=4\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)$
$\Rightarrow b^{2}\left(a^{2}+c^{2}+2 a c\right)=a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2}+b^{4}$
$\Rightarrow a^{2} b^{2}+b^{2} c^{2}+2 a c b^{2}=a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2}+b^{4}$
$\Rightarrow b^{4}+a^{2} c^{2}-2 a c b^{2}=0$
$\Rightarrow\left(b^{2}-a c\right)^{2}=0 \Rightarrow b^{2}=a c$
therefore, discriminant $=0$
$\Rightarrow[2 b(a+c)]^{2}-4\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)=0$
$4 b^{2}(a+c)^{2}=4\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)$
$\Rightarrow b^{2}\left(a^{2}+c^{2}+2 a c\right)=a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2}+b^{4}$
$\Rightarrow a^{2} b^{2}+b^{2} c^{2}+2 a c b^{2}=a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2}+b^{4}$
$\Rightarrow b^{4}+a^{2} c^{2}-2 a c b^{2}=0$
$\Rightarrow\left(b^{2}-a c\right)^{2}=0 \Rightarrow b^{2}=a c$
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