JEE Main 2025 Jan 23 Shift 1, Mathematics Q25: Quadratic Equations

JEE Main2025Jan 23, Shift 1Mathematics

If the equation $a (b - c) x^{2} + b (c - a) x + c (a - b) = 0$ has equal roots, where $a + c = 15$ and $b = \frac{36}{5}$ , then $a^{2} + c^{2}$ is equal to ________

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Solution

Note that the coefficients sum to $a (b - c) + b (c - a) + c (a - b) = 0$ , so $x = 1$ is always a root. For equal roots, the other root must also be $1$ , giving product of roots $= 1$ :

$\frac{c ( a - b )}{a ( b - c )} = 1 \Rightarrow c (a - b) = a (b - c) \Rightarrow a c - b c = ab - a c \Rightarrow 2 a c = b (a + c)$ .

Substitute $a + c = 15$ and $b = 36/5$ : $2 a c = \frac{36}{5} \cdot 15 = 108 \Rightarrow a c = 54$ .

$a^{2} + c^{2} = (a + c)^{2} - 2 a c = 225 - 108 = 117$ .

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JEE Main 2025 Jan 23 Shift 1, Mathematics Q25: Quadratic Equations

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