If the roots of the equation $(a^2+b^2)x^2-2(ac+bd)x+(c^2+d^2)=0$ are equal, prove that $\frac{a}{b}=\frac{c}{d}$.

Given:

Given quadratic equation is $(a^2+b^2)x^2-2(ac+bd)x+(c^2+d^2)=0$. The roots of the given quadratic equation are equal.

To do:

We have to prove that $\frac{a}{b}=\frac{c}{d}$.

Solution:

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

$a=(a^2+b^2), b=-2(ac+bd)$ and $c=(c^2+d^2)$.

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

$D=[-2(ac+bd)]^2-4(a^2+b^2)(c^2+d^2)$

$D=4(a^2c^2+2abcd+b^2d^2)-4(a^2c^2+a^2d^2+b^2c^2+b^2d^2)$

$D=4a^2c^2+8abcd+4b^2d^2-4a^2c^2-4a^2d^2-4b^2c^2-4b^2d^2$

$D=8abcd-4a^2d^2-4b^2c^2$

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

This implies,

$8abcd-4a^2d^2-4b^2c^2=0$

$4a^2d^2+4b^2c^2-8abcd=0$

$(2ad-2bc)^2=0$

$2ad-2bc=0$

$2ad=2bc$

$ad=bc$

$\frac{a}{b}=\frac{c}{d}$

Hence proved.

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Updated on: 10-Oct-2022

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