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

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Solution

For an equation the nature of the roots can be determined by the value of D.
where D =

If, D=0 , real and equal roots
D=0

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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

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

a d = b c

(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}

(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}

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