Question
If the equation $\left(1+m^2\right) x^2+2 m c x+\left(c^2-a^2\right)=0$ has equal roots, prove that $c^2=a^2\left(1+m^2\right)$.
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Answer
The given equation $\left(1+m^2\right) x^2+2 m c x+\left(c^2-a^2\right)=0$, has equal roots,
Then prove that $c ^2= a ^2\left(1+ m ^2\right)$
Here, $a=\left(1+m^2\right), b=2 m c$ and $c=\left(c^2-a^2\right)$
As we know that $D=b^2-4 a c$
Putting the value of $a=\left(1+m^2\right), b=2 m c$ and $c=\left(c^2-a^2\right)$
$\Rightarrow D=b^2-4 a c$
$\Rightarrow D=\{2 m c\}^2-4 \times\left(1+m^2\right) \times\left(c^2-a^2\right)$
$\Rightarrow D=4\left(m^2 c^2\right)-4\left(c^2-a^2+m^2 c^2-m^2 a^2\right)$
$\Rightarrow D=4 m^2 c^2-4 c^2+4 a^2-4 m^2 c^2+4 m^2 a^2$
$\Rightarrow D=4 a^2+4 m^2 a^2-4 c^2$
The given equation will have real roots, if $D=0$
$\Rightarrow 4 a^2+4 m^2 a^2-4 c^2=0$
$\Rightarrow 4 a^2+4 m^2 a^2=4 c^2$
$\Rightarrow 4 a^2\left(1+m^2\right)=4 c^2$
Hence, $c^2=a^2\left(1+m^2\right)$
Then prove that $c ^2= a ^2\left(1+ m ^2\right)$
Here, $a=\left(1+m^2\right), b=2 m c$ and $c=\left(c^2-a^2\right)$
As we know that $D=b^2-4 a c$
Putting the value of $a=\left(1+m^2\right), b=2 m c$ and $c=\left(c^2-a^2\right)$
$\Rightarrow D=b^2-4 a c$
$\Rightarrow D=\{2 m c\}^2-4 \times\left(1+m^2\right) \times\left(c^2-a^2\right)$
$\Rightarrow D=4\left(m^2 c^2\right)-4\left(c^2-a^2+m^2 c^2-m^2 a^2\right)$
$\Rightarrow D=4 m^2 c^2-4 c^2+4 a^2-4 m^2 c^2+4 m^2 a^2$
$\Rightarrow D=4 a^2+4 m^2 a^2-4 c^2$
The given equation will have real roots, if $D=0$
$\Rightarrow 4 a^2+4 m^2 a^2-4 c^2=0$
$\Rightarrow 4 a^2+4 m^2 a^2=4 c^2$
$\Rightarrow 4 a^2\left(1+m^2\right)=4 c^2$
Hence, $c^2=a^2\left(1+m^2\right)$
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