2015/060) If the roots of the equation

$(a²+b²)x²-2(ac+bd)x+(c²+d²)=0$ are equal, prove that $\dfrac{a}{b}=\dfrac{c}{d}$

Solution

For roots to be equal, discriminant = 0
$=> [-2(ac + bd)]^2 - 4 * (a^2 + b^2) (c^2 + d^2) = 0$
$=> 4[(ac)^2 + 2abcd + (cd)^2] - 4[(ac)^2 + (ad)^2 + (bc)^2 + (cd)^2] = 0$
$=> [(ac)^2 + 2abcd + (cd)^2] - [(ac)^2 + (ad)^2 + (bc)^2 + (cd)^2] = 0$
$=> (2abcd - (bc)^2 - (ad)^2) = 0$
$=> (bc)^2 - 2abcd + (ad)^2 = 0$
$=> (bc-ad)^2 = 0$
$=> bc = ad$
$=> \dfrac{a}{b} = \dfrac{c}{d}$