The number of distinct real solutions of the equation x x +4 +3 x +2 +10=0 is

Mathematics · Quadratic Equations

JEE Main 2026 — 22 January, Morning Shift — Question 8

The number of distinct real solutions of the equation $\mathrm{x}|\mathrm{x}+4|+3|\mathrm{x}+2|+10=0$ is

Case I $\mathrm{x}<-4$

$x(-(x+4))+3(-(x+2))+10=0$

$\mathrm{x}^{2}+7 \mathrm{x}-4=0$

$\Rightarrow x=-\frac{7+\sqrt{65}}{2}$ or $-\frac{7-\sqrt{65}}{2}$

Case II $-4 \leq \mathrm{x}<-2$

$x(x+4)+3(-(x+2))+10=0$

$\mathrm{x}^{2}+\mathrm{x}+4=0$

$D < 0$ No solution

Case III $\mathrm{x} \geq-1$

$x(x+4)+3(x+2)+10=0$

$\mathrm{x}^{2}+7 \mathrm{x}+16=0$

$\mathrm{D}<0$

No solution ⇒ No. of solution $=1$ .

Answer key and solution verified before publishing.

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