Algebra – Quadratic Equations – JEE Main 22 Jan 2026 Shift 1

Solution:

We analyze the equation in different intervals based on the critical points of the absolute value functions: $x = -4$ and $x = -2$.

**Case 1:** $x < -4$ Here, $|x+4| = -(x+4)$ and $|x+2| = -(x+2)$. The equation becomes: $$ x(-(x+4)) + 3(-(x+2)) + 10 = 0 $$ $$ -x^2 - 4x - 3x - 6 + 10 = 0 $$ $$ -x^2 - 7x + 4 = 0 \Rightarrow x^2 + 7x - 4 = 0 $$ $$ x = \frac{-7 \pm \sqrt{49 - 4(1)(-4)}}{2} = \frac{-7 \pm \sqrt{65}}{2} $$ Since $\sqrt{65} \approx 8.06$: $x_1 = \frac{-7 - 8.06}{2} \approx -7.53$ (Accepted as $x < -4$) $x_2 = \frac{-7 + 8.06}{2} \approx 0.53$ (Rejected as $x \not< -4$) So, 1 solution in this interval.
**Case 2:** $-4 \le x < -2$ Here, $|x+4| = (x+4)$ and $|x+2| = -(x+2)$. The equation becomes: $$ x(x+4) + 3(-(x+2)) + 10 = 0 $$ $$ x^2 + 4x - 3x - 6 + 10 = 0 $$ $$ x^2 + x + 4 = 0 $$ Discriminant $D = 1^2 - 4(1)(4) = 1 - 16 = -15 < 0$. No real solution in this interval.
**Case 3:** $x \ge -2$
Here, $|x+4| = (x+4)$ and $|x+2| = (x+2)$.
The equation becomes:
$$ x(x+4) + 3(x+2) + 10 = 0 $$
$$ x^2 + 4x + 3x + 6 + 10 = 0 $$
$$ x^2 + 7x + 16 = 0 $$
Discriminant $D = 7^2 – 4(1)(16) = 49 – 64 = -15 < 0$. No real solution in this interval.
Total number of distinct real solutions is 1.

Ans. (2)