JEE Main 2025 Jan 28 Shift 1, Mathematics Q15: Some Other Forms of Equations

JEE Main2025Jan 28, Shift 1Mathematics

The sum of the squares of all the roots of the equation $x^{2} + ∣2 x - 3∣ - 4 = 0$ , is :

A
$3 (3 - 2)$
B
$6 (3 - 2)$
C
$6 (2 - 2)$
D
$3 (2 - 2)$

Solution

Split on the sign of $2 x - 3$ .

Case 1: $x \geq \frac{3}{2}$ . Equation becomes $x^{2} + 2 x - 7 = 0$ , with roots $x = - 1 \pm 22$ . Only $x = 22 - 1 \approx 1.83$ satisfies $x \geq \frac{3}{2}$ . The other root is negative and rejected.

Case 2: $x < \frac{3}{2}$ . Equation becomes $x^{2} - 2 x - 1 = 0$ , with roots $x = 1 \pm 2$ . Both must be checked: $1 + 2 \approx 2.41 \geq \frac{3}{2}$ , rejected; $1 - 2 \approx - 0.41 < \frac{3}{2}$ , accepted.

So the valid roots are $22 - 1$ and $1 - 2$ .

Sum of squares: $(22 - 1)^{2} + (1 - 2)^{2} = (8 - 42 + 1) + (1 - 22 + 2) = 12 - 62 = 6 (2 - 2)$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q15: Some Other Forms of Equations

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