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

Open in App

Solution

$(a ² + b ²) x ² - 2 (a b + c d) x + (c ² + d ²) = 0$ roots are equal so,

$D = b ² - 4 a c = 0$

$2 (a b + c d) ² - 4 (a ² + b ²) (c ² + d ²) = 0$

$4 (a b + c d) ² - 4 (a ² + b ²) (c ² + d ²) = 0$

$(a ² b ² + c ² d ² + 2 a b c d) - a ² c ² - a ² d ² - b ² d ² - b ² c ² = 0$

$- a ² c ² - b ² d ² + 2 a b c d = 0$

$- (a ² c ² + b ² d ² - 2 a b c d) = 0$

$(a c - b d) ² = 0$

$a c - b d = 0$

$a c = b d$

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

Suggest Corrections

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

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

a d \neq b c

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

Join BYJU'S Learning Program