If the equations (a2+b2)x2−2(ac+bd)x+c2+d2=0 has equal roots, then

Solution:

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

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

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, $c=c^{2}+d^{2}$

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

$=4 a^{2} c^{2}+4 b^{2} d^{2}+8 a b c d-4\left(a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2}\right)$

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

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

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

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

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

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

$a d-b c=0$

$a d=b c$

Thus, the correct answer is (b)