First slide

Theory of equations

Question

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

Moderate

A

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

$\frac{a}{c} + \frac{b}{d} = 0$
C

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

a + b =a + d

Solution

Since, roots are real.
$\begin{array}{l} ∴ {2 (b c + a d)}^{2} = 4 (a^{2} + b^{2}) (c^{2} + d^{2}) \\ \Rightarrow 4 b^{2} c^{2} + 4 a^{2} d^{2} + 8 a b c d = 4 a^{2} c^{2} + 4 a^{2} d^{2} + 4 b^{2} c^{2} + 4 b^{2} d^{2} \\ \Rightarrow 4 a^{2} d^{2} + 4 b^{2} c^{2} - 8 a b c d = 0 \\ \Rightarrow 4 (a d - b c)^{2} = 0 \\ \Rightarrow a d = b c \\ \Rightarrow \frac{a}{b} = \frac{c}{d} \end{array}$

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