First slide

Theory of equations

Question

$For a, b, c non-zero, real distinct, the equation, (a^{2} + b^{2}) x^{2} - 2 b (a + c) x + b^{2} + c^{2} = 0 has non-zero real roots . One of these roots$ $is also the root of the equation:$

Moderate

A

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

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

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

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

Solution

$\begin{array}{l} (a^{2} + b^{2}) x^{2} - 2 b (a + c) x + b^{2} + c^{2} = 0 \\ D = 4 b^{2} (a + c)^{2} - 4 (a^{2} + b^{2}) (b^{2} + c^{2}) \\ = - 4 (b^{4} - 2 b^{2} a c + a^{2} c^{2}) \end{array}$
$= - 4 {(b^{2} - a c)}^{2}$
$For real roots, D \geq 0$
$\begin{array}{l} \Rightarrow - 4 {(b^{2} - a c)}^{2} \geq 0 \\ \Rightarrow b^{2} - a c = 0 \\ \Rightarrow Roots are real and equal. \end{array}$
$\begin{array}{l} ∴ Roots are \frac{2 b (a + c) \pm 0}{2 (a^{2} + b^{2})} \\ = \frac{b (a + c)}{a^{2} + a c} \\ = \frac{b}{a} \end{array}$
This root satisfies option (c).

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