If equation has $(a^{2} + b^{2}) x^{2} - 2 (a c + b d) x + (c^{2} + d^{2}) = 0$ equal roots , then prove that $\frac{a}{b} = \frac{c}{d}$

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Solution

If equation has equal root then

$b^{2} - 4 a c = 0$

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

$\Rightarrow 4 (a^{2} c^{2} + b^{2} + 2 a b c d)$

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

$\Rightarrow a^{2} c^{2} + b^{2} d^{2} + 2 a b c d d = a^{2} c^{2} + b^{2} c^{2} + a^{2} d^{2} + b^{2} d^{2}$

$\Rightarrow 2 a b c d = b^{2} c^{2} + a^{2} d^{2}$

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

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

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

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

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Similar questions

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

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

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

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

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

\neq

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

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

a d = \sqrt{b c}

a b = \sqrt{c d}

(a^{2} + b^{2}) x 2 - 2 (a c + b d) x + (c^{2} + d^{2}) = 0 a r e e q u a l, p r o v e t h a t \frac{a}{b} = \frac{c}{d}

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