If the roots of the equation $(a^{2} + b^{2}) x^{2} - 2 (a c + b d) x + (c^{2} + b^{2}) = 0$ are equal then prove that $a d = b c$

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Solution

The discriminant of the given equation is given by
$D = [- 2 (a c + b d)]^{2} - 4 \times (a^{2} + b^{2}) \times (c^{2} + d^{2})$
$D = 4 [a c + b d]^{2} - 4 \times (a^{2} c^{2} + a^{2} d^{2} + b^{2} c^{2} + b^{2} d^{2})$
$D = 4 [a^{2} c^{2} + b^{2} d^{2} + 2 a b c d] - 4 \times (a^{2} c^{2} + a^{2} d^{2} + b^{2} c^{2} + b^{2} d^{2})$
$D = - 4 {(a d - b c)}^{2}$
The given equation will have equal roots, if
$D = 0$
$- 4 {(a d - b c)}^{2} = 0$
$a d = b c$

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