Q48 / 120 · NDA 2024 (Phase 2) · Mathematics

Paragraph: The roots of the quadratic equation $a^2\left(b^2-c^2\right) x^2+b^2\left(c^2-a^2\right) x+c^2\left(a^2-b^2\right)=0$ are equal $\left(a^2 \neq b^2 \neq c^2\right)$.
Question: Which one of the following is a root of the equation?

  • A
    $\frac{b^2\left(c^2-a^2\right)}{a^2\left(c^2-b^2\right)}$
  • B
    $\frac{b^2\left(c^2-a^2\right)}{a^2\left(b^2-c^2\right)}$
  • Correct answer
    C
    $\frac{b^2\left(c^2-a^2\right)}{2 a^2\left(c^2-b^2\right)}$
  • D
    $\frac{b^2\left(c^2-a^2\right)}{2 a^2\left(b^2-c^2\right)}$

Solution

The given equation is as follows, $a^2\left(b^2-c^2\right) x^2+b^2\left(c^2-a^2\right) x+c^2\left(a^2-b^2\right)=0$ ....(i) Since, equation has equal roots $\Rightarrow D=0$ $\begin{aligned} & \Rightarrow \quad\left[b^2\left(c^2-a^2\right)\right]^2-4 a^2 c^2\left(b^2-c^2\right)\left(a^2+b^2\right)=0 \\ & \Rightarrow \quad b^4\left(c^2-a^2\right)^2-4 a^2 c^2\left(b^2-c^2\right)\left(a^2-b^2\right)=0 \end{aligned}$ Solving above equation, we have, $\begin{aligned} & b^4\left(c^2+a^2\right)^2=4 a^4 c^4 \Rightarrow b^2\left(a^2+c^2\right)^2=2 a^2 c^2 \\ & \Rightarrow \frac{2}{b^2}=\frac{1}{a^2}+\frac{1}{c^2} \end{aligned}$ $\therefore a^2, b^2, c^2$ and in H.P. Let $\alpha$ be the required root of the given equation. Now, Sum of roots $=-\frac{B}{A}=-\frac{b^2\left(c^2-a^2\right)}{a^2\left(b^2-c^2\right)}$ $\Rightarrow \alpha+\alpha=\frac{-b^2\left(c^2-a^2\right)}{a^2\left(b^2-c^2\right)} \Rightarrow \alpha=\frac{b^2\left(c^2-a^2\right)}{2 a^2\left(c^2-b^2\right) a}$

Hence, the correct answer is option: 'C'.

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