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UPSC NDA II 2024 Mathematics Paper

Question Numbers: 47-48

Direction : Consider the following for the items that follow :

The roots of the quadratic equation

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

are equal (

a^{2} \neq b^{2} \neq c^{2}

Question : 48 of 120

Marks: +1, -0

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

Solution:

Concept:

The given equation has equal roots. For a quadratic

Ax^2+Bx+C=0

, if both roots are equal to

\alpha

, then

\alpha+\alpha = -\frac{B}{A}

Explanation:

Let the required root be

\alpha

Since the roots are equal, sum of roots

= \alpha + \alpha = 2\alpha

Given that sum of roots

= -\frac{B}{A} = \frac{-b^2(c^2-a^2)}{a^2(b^2-c^2)}

Thus,

2\alpha = \frac{-b^2(c^2-a^2)}{a^2(b^2-c^2)}

Divide both sides by

2

\alpha = \frac{-b^2(c^2-a^2)}{2a^2(b^2-c^2)}

Simplify:

\alpha = \frac{b^2(c^2-a^2)}{2a^2(c^2-b^2)}

because

-(b^2-c^2) = c^2-b^2

This matches option C.

Answer:

Option C

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