If the equation (1+m^2 ) x^2+2 m c x+ (c^2-a^2 )=0 has equal roots, prove that c^2=a^2 (1+m^2 ).

The given equation (1+m^2 ) x^2+2 m c x+ (c^2-a^2 )=0, has equal roots, Then prove that c^2=a^2 (1+m^2 ) Here, a= (1+m^2 ), b=2 m c and c= (c^2-a^2 ) As we know that D=b^2-4 a c Putting the value of a= (1+m^2 ), b=2 m c and c= (c^2-a^2 ) D = b^2 - 4ac D = 2mc ^2 - 4 (1 + m^2) (c^2 - a^2) D = 4(m^2c^2) - 4(c^2 - a^2 + m^2c^2 - m^2a^2) D = 4m^2c^2 - 4c^2 + 4a^2 - 4m^2c^2 + 4m^2a^2 D = 4a^2 + 4m^2a^2 - 4c^2 The given equation will have real roots, if D = 0 4a^2 + 4m^2a^2 - 4c^2 = 0 4a^2 + 4m^2a^2 = 4c^2 4a^2(1 + m^2) = 4c^2 Hence, c^2 = a^2(1 + m^2)

If the equation (1+m^2 ) x^2+2 m c x+ (c^2-a^2 )=0 has equal root…

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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 $c=\left(c^2-a^2\right)$
$\Rightarrow D = b^2 - 4ac$
$\Rightarrow D = {2mc}^2 - 4 \times (1 + m^2) \times (c^2 - a^2)$
$\Rightarrow D = 4(m^2c^2) - 4(c^2 - a^2 + m^2c^2 - m^2a^2)$
$\Rightarrow D = 4m^2c^2 - 4c^2 + 4a^2 - 4m^2c^2 + 4m^2a^2$
$\Rightarrow D = 4a^2 + 4m^2a^2 - 4c^2$
The given equation will have real roots, if D = 0
$\Rightarrow 4a^2 + 4m^2a^2 - 4c^2 = 0$
$\Rightarrow 4a^2 + 4m^2a^2 = 4c^2$
$\Rightarrow 4a^2(1 + m^2) = 4c^2$
$Hence, c^2 = a^2(1 + m^2)$