Aspire Faculty ID #18055 · Topic: JEE Main 2026 (24 January Morning Shift) · Just now
JEE Main 2026 (24 January Morning Shift)

The number of the real solutions of the equation: $x|x+3|+|x-1|-2=0$ is:

Solution

Break into intervals

Case I: $x\ge1$

$x(x+3)+(x-1)-2=0$

$x^2+4x-3=0$

$x=-2\pm\sqrt{7}$ (reject)

Case II: $-3\le x<1$

$x(x+3)+(1-x)-2=0$

$x^2+2x-1=0$

$x=-1\pm\sqrt{2}$

Case III: $x<-3$

$x(-(x+3))+(1-x)-2=0$

$x^2+4x+1=0$

$x=-2\pm\sqrt{3}$ (reject)

Only valid solutions = $3$

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