If ad $\neq$ bc, then prove that the equation $(a^{2} + b^{2}) x^{2} + 2 (a c + b d) x + (c^{2} + d^{2}) = 0$ has no real roots.

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Solution

Given, ad $\neq$ bc

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

D = $B^{2}$ - 4AC = ${[2 (a c + b d)]}^{2} - 4 (a^{2} + b^{2}) (c^{2} + d^{2})$

= $4 [a^{2} c^{2} + b^{2} d^{2} + 2 a b c d] - 4 (a^{2} c^{2} + a^{2} d^{2} + b^{2} c^{2} + b^{2} d^{2})$

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

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

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

= $- 4 {[a d - b c]}^{2}$

As D is negative

Hence, given equation has no real roots.
Hence proved

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