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

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Solution

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

Comparing with $A x^{2} + B x + c = 0$

$A = (a^{2} + b^{2})$

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

$C = c^{2} + d^{2}$

$b^{2} - 4 a c$

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

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

$= 8 a b c d - 4 (a^{2} d^{2} + b^{2} c^{2})$

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

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

Roots are equal if ad = bc

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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

(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 = \sqrt{b c}

a b = \sqrt{c d}

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