Prove that the equation $x^{2} (a^{2} + b^{2}) + 2 x (a c + b d) + (c^{2} + d^{2}) = 0$ has no real root, if $a d \neq b c$ .

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Solution

Discriminant,
$D = 4 (a c + b d)^{2} - 4 (a^{2} + b^{2}) (c^{2} + d^{2})$
$D = 4 [(a c + b d)^{2} - (a^{2} + b^{2}) (c^{2} + d^{2})]$
$D = 4 [a^{2} c^{2} + b^{2} d^{2} + 2 a c b d - a^{2} c^{2} - a^{2} d^{2} - b^{2} c^{2} - b^{2} d^{2}]$
$D = 4 [2 a c b d - a^{2} d^{2} - b^{2} c^{2}]$
$D = - 4 [a^{2} d^{2} + b^{2} c^{2} - 2 a d b c]$
$D = - 4 (a d - b c)^{2}$
We have, $a d \neq b c$
Therefore,
$a d - b c \neq 0$
$(a d - b c)^{2} > 0$
$- 4 (a d - b c)^{2} < 0$
$D < 0$
Hence, the given equation has no real roots.

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