Evaluate $c^2 -a^2(1 + m^2)$ if roots are equal for the quadratic equation $(1+m^2)x^2$ $+$ $2cmx$ $+$ $(c^2 \,–\, a^2) = 0$

$(1+m^2)x^2 + 2cmx + (c^2 \,–\, a^2) = 0$ is a quadratic equation and the roots of this equation are same.

01

Rule of Equal Roots of Quadratic equation

Due to the equal roots of the quadratic equation, the discriminant of the quadratic equation is zero. For example, $ax^2+bx+c = 0$ is a quadratic equation in general form. The discriminant is $b^2-4ac$ and it is zero due to equal roots.

$b^2 \,-\, 4ac = 0$

02

Evaluation of Discriminant

Compare the quadratic equation $(1+m^2)x^2 + 2cmx + (c^2 \,–\, a^2) = 0$ with standard form quadratic equation $ax^2+bx+c = 0$.

$a = 1+m^2$, $b = 2cm$ and $c = (c^2 \,–\, a^2)$. Substitute these values in discriminate of the quadratic equation.

$b^2 \,-\, 4ac = 0$

$\implies {(2cm)}^2 \,-\, 4 \times (1+m^2) \times (c^2 \,–\, a^2) = 0$

$\implies 4c^2m^2 \,-\, 4 \times (c^2 \,–\, a^2 + m^2 c^2 \,–\, m^2 a^2) = 0$

$\implies 4c^2m^2 \,-\, 4c^2 + 4a^2 \,-\, 4m^2c^2 + 4m^2a^2 = 0$

$\implies 4c^2m^2 \,-\, 4c^2 + 4a^2 \,-\, 4c^2m^2 + 4m^2a^2 = 0$

$\require{cancel} \implies \cancel{4c^2m^2} \,-\, 4c^2 + 4a^2 \,-\, \cancel{4c^2m^2} + 4m^2a^2 = 0$

$\implies \,-\, 4c^2 + 4a^2 + 4m^2a^2 = 0$

$\implies 4a^2 + 4m^2a^2 = 4c^2$

$\implies 4c^2 = 4a^2 + 4m^2a^2$

$\implies 4c^2 = 4a^2(1 + m^2)$

$\require{cancel} \implies \cancel{4}c^2 = \cancel{4}a^2(1 + m^2)$

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

$\therefore \,\,\,\,\,\, c^2 -a^2(1 + m^2) = 0$

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