Q47 / 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 statements is correct?

  • A
    $a^2, b^2, c^2$ are in AP.
  • B
    $a^2, b^2, c^2$ are in GP.
  • Correct answer
    C
    $a^2, b^2, c^2$ are in HP.
  • D
    $a^2, b^2, c^2$ are neither in AP nor in GP nor in HP.

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. Option (c) is the correct option.

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

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