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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2024 · Mathematics · Quadratic Equation

Paragraph: The roots of the quadratic equation a^2(b^2-c^2) x^2+b^2...

Paragraph: The roots of the quadratic equation a2(b2−c2)x2+b2a^2\left(b^2-c^2\right) x^2+b^2\left(c²-a²\right) x+c² \left(a^2-b²\right)=0areequal(a2≠b2≠c2)are equal \left(a^2 ≠ b² ≠ c²\right).
Question: Which one of the following is a root of the equation?

Options

  1. A\frac{b^2\left(c^2-a^2\right)}{a^2\left(c^2-b^2\right)}
  2. B\frac{b^2\left(c^2-a^2\right)}{a^2\left(b^2-c^2\right)}
  3. C\frac{b^2\left(c^2-a^2\right)}{2 a^2\left(c^2-b^2\right)}
  4. D\frac{b^2\left(c^2-a^2\right)}{2 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)}

Step-by-step solution

The given equation is as follows, a2a^2\left(b^2-c^2\right) x^2+b^2(c2−a2)x+c2(a2−b2)=0\left(c^2-a^2\right) x+c^2\left(a^2-b^2\right)=0 ....(i) Since, equation has equal roots ⇒D=0\Rightarrow D=0⇒[b2(c2−a2)]2−4a2c2(b2−c2)(a2+b2)=0⇒b4(c2−a2)2−4a2c2(b2−c2)(a2−b2)=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}Solvingaboveequation,wehave,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}$∴a2,b2,c2\therefore a^2, b^2, c^2 and in H.P. Let α\alpha be the required root of the given equation. Now, Sum of roots =−BA=−b2(c2−a2)a2(b2−c2)⇒α+α=−b2(c2−a2)a2(b2−c2)⇒α=b2(c2−a2)2a2(c2−b2)a=-\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}
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