If the roots of the equation (a2 + b2)x2 − 2 (ac + bd)x + (c2 + d2) = 0

The given quadric equation is $\left(a^{2}+b^{2}\right) x^{2}-2(a c+b d) x+\left(c^{2}+d^{2}\right)=0$, and roots are real

Then prove that $\frac{a}{b}=\frac{c}{d}$.

Here,

$a=\left(a^{2}+b^{2}\right), b=-2(a c+b d)$ and, $c=\left(c^{2}+d^{2}\right)$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=\left(a^{2}+b^{2}\right), b=-2(a c+b d)$ and, $c=\left(c^{2}+d^{2}\right)$

$D=b^{2}-4 a c$

$=\{-2(a c+b d)\}^{2}-4 \times\left(a^{2}+b^{2}\right) \times\left(c^{2}+d^{2}\right)$

$=4\left(a^{2} c^{2}+2 a b c d+b^{2} d^{2}\right)-4\left(a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2}\right)$

$=-4 a^{2} d^{2}-4 b^{2} c^{2}+8 a b c d$

$=-4\left(a^{2} d^{2}+b^{2} c^{2}-2 a b c d\right)$

The given equation will have real roots, if $D=0$

$-4\left(a^{2} d^{2}+b^{2} c^{2}-2 a b c d\right)=0$

$\left(a^{2} d^{2}+b^{2} c^{2}-2 a b c d\right)=0$

$(a d)^{2}+(b c)^{2}-2(a d)(b c)=0$

$(a d-b c)^{2}=0$

Square root both sides we get,

$a d-b c=0$

$a d=b c$

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

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