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, then prove that $\frac{a}{b} = \frac{c}{d}$ .

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Solution

$(a^{2} + b^{2}) x^{2} - 2 (a c + b d) x + (c^{2} + d^{2}) = 0$
Comparing with $A x^{2} + B x + C = 0$
$A = (a^{2} + b^{2}), B = - 2 (a c + b d), C = (c^{2} + d^{2})$
For equal roots, $B^{2} - 4 A C = 0$
$(- 2 (a c + b d))^{2} - 4 (a^{2} + b^{2}) (c^{2} + d^{2}) = 0$
$4 (a^{2} c^{2} + 2 a b c d + b^{2} d^{2}) - 4 (a^{2} c^{2} + a^{2} d^{2} + b^{2} c^{2} + b^{2} d^{2}) = 0$ , divide by 4 to get
$a^{2} c^{2} + 2 a b c d + b^{2} d^{2} - a^{2} c^{2} - a^{2} d^{2} - b^{2} c^{2} - b^{2} d^{2} = 0$
$2 a b c d - a^{2} d^{2} - b^{2} c^{2} = 0$ , Multiply by -1 to get,
$a^{2} d^{2} - 2 a b c d + b^{2} c^{2} = 0$
$(a d - b c)^{2} = 0$ , Take square root on both sides to get,
$a d - b c = 0$
$a d = b c$
$\frac{a}{b} = \frac{c}{d}$

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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

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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}

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