(B) We analyze the equation $x|x+4|+3|x+2|+10=0$ by considering different intervals for $x$: Case $I$: $x < -4$. The equation becomes $x(-(x+4)) + 3(-

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(B) We analyze the equation $x|x+4|+3|x+2|+10=0$ by considering different intervals for $x$: Case $I$: $x Case $II$: $-4 \leq x Case $III$: $x \geq -2$. The equation becomes $x(x+4) + 3(x+2) + 10 = 0$, which simplifies to $x^2 + 4x + 3x + 6 + 10 = 0$, or $x^2 + 7x + 16 = 0$. The discriminant $D = 7^2 - 4(1)(16) = 49 - 64 = -15 Thus, there is only $1$ distinct real solution.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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

JEE MAIN 2026 All JEE MAIN

A
$3$
B
$1$
C
$0$
D
$2$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Solution of quadratic equations and Nature of roots

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