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

Given:

Given quadratic equation is $(1+m^2)x^2+2mcx+(c^2-a^2)=0$. The roots of the given quadratic equation are equal.

To do:

We have to prove that $c^2=a^2(1+m^2)$.

Solution:

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=(1+m^2), b=2mc$ and $c=(c^2-a^2)$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

$D=(2mc)^2-4(1+m^2)(c^2-a^2)$

$D=4m^2c^2-4(c^2-a^2+m^2c^2-m^2a^2)$

$D=4(m^2c^2-c^2+a^2-m^2c^2+m^2a^2)$

$D=4(m^2a^2+a^2-c^2)$

The given quadratic equation has equal roots if $D=0$.

This implies,

$4(m^2a^2+a^2-c^2)=0$

$m^2a^2+a^2-c^2=0$

$a^2(m^2+1)=c^2$

$c^2=a^2(1+m^2)$

Hence proved.

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Updated on: 10-Oct-2022

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