Q. If the roots of $\left( a^{2} + b^{2} \right)x^{2} - 2(bc + ad)x + c^{2} + d^{2} = 0$ are equal, then
A) $\frac{a}{b} = \frac{c}{d}$
B) $\frac{a}{c} + \frac{b}{d} = 0$
C) $\frac{a}{d} = \frac{b}{c}$
D) $a + b = c + d$
Explanation
(a)
Since roots are real.
$\therefore\left\{ 2(bc + ad) \right\}^{2} = 4\left( a^{2} + b^{2} \right)(c^{2} + d^{2})$
$\Rightarrow 4b^{2}c^{2} + 4a^{2}d^{2} + 8abcd = 4a^{2}c^{2} + 4a^{2}d^{2} + 4b^{2}c^{2} + 4b^{2}d^{2}$
$\Rightarrow \ 4a^{2}d^{2} + 4b^{2}c^{2} - 8abcd = 0\$
$\Rightarrow 4(ad - bc)^{2} = 0$
$\Rightarrow ad = bc$
$\Rightarrow \frac{a}{b} = \frac{c}{d}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.