Cheek whether the quadratic equation $(a^{2} + b^{2}) x^{2} + 2 (a c + b d) x + c^{2} + d^{2} = 0$ has real roots.

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Solution

$(a^{2} + b^{2}) x^{2} + 2 (a c + b d) x + (c^{2} + d^{2}) = 0$
$D = b^{2} - 4 a c$
$D = 4 (a c + b d)^{2} - 4 (a^{2} + b^{2}) (c^{2} + d^{2})$
$D = 4 (a^{2} c^{2} + b^{2} d^{2} + 2 a b c d) - 4 (a^{2} c^{2} + a^{2} d^{2} + b^{2} c^{2} + b^{2} d^{2})$
$D = 4 (2 a b c d - a^{2} d^{2} - b^{2} c^{2})$
$D = - 4 (a d - b c)^{2}$
Since $(a d - b c)^{2}$ is always positive.
$∴$ Roots are not real.
They are imaginary.

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Similar questions

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

a d \neq b c

(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

(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}

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

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

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