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NDA 2 2024 Maths, question 48

Topic: Algebra

Passage · Q47 to Q48

The roots of the quadratic equation

a2(b2−c2)x2+b2(c2−a2)x+c2(a2−b2)=0a^{2}(b^{2}-c^{2})x^{2}+b^{2}(c^{2}-a^{2})x+c^{2}(a^{2}-b^{2})=0

are equal (a2≠b2≠c2)(a^{2}\ne b^{2}\ne c^{2}).

Q48

Which one of the following is a root of the equation?

  1. (a) b2(c2−a2)a2(c2−b2)\frac{b^{2}(c^{2}-a^{2})}{a^{2}(c^{2}-b^{2})}
  2. (b) b2(c2−a2)a2(b2−c2)\frac{b^{2}(c^{2}-a^{2})}{a^{2}(b^{2}-c^{2})}
  3. (c) b2(c2−a2)2a2(c2−b2)\frac{b^{2}(c^{2}-a^{2})}{2a^{2}(c^{2}-b^{2})}Correct answer
  4. (d) b2(c2−a2)2a2(b2−c2)\frac{b^{2}(c^{2}-a^{2})}{2a^{2}(b^{2}-c^{2})}

Solution.

Answer: (c) b2(c2−a2)2a2(c2−b2)\frac{b^{2}(c^{2}-a^{2})}{2a^{2}(c^{2}-b^{2})}

  1. For Ax²+Bx+C=0 with equal roots, the repeated root is −B/(2A).
  2. Here this equals b²(c²−a²)/[2a²(c²−b²)], option (c).
  3. It also equals 1, as checked by summing the original coefficients.