If $a d \neq b c$ , the prove that the equation $(a^{2} + b^{2}) x^{2} + 2 (a c + b d) x + (c^{2} + d^{2}) = 0$ has no real roots.

Open in App

Solution

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

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

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

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

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

$i f a d \neq b c$

$D < 0$
$\Rightarrow$ it has no real roots.

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

(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 d = \sqrt{b c}

a b = \sqrt{c d}

Join BYJU'S Learning Program