The number of real solutions of the equation | vedclass

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(A) To find the number of real solutions for $x|x+3|+|x-1|-2=0$, we analyze the equation in three intervals based on the critical points $x=-3$ and $x

Vedclass · Accepted Answer

$3$

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(A) To find the number of real solutions for $x|x+3|+|x-1|-2=0$, we analyze the equation in three intervals based on the critical points $x=-3$ and $x=1$: $I$. Case $x \ge 1$: $x(x+3) + (x-1) - 2 = 0 \implies x^2+3x+x-1-2=0 \implies x^2+4x-3=0$. Using the quadratic formula, $x = \frac{-4 \pm \sqrt{16 - 4(1)(-3)}}{2} = -2 \pm \sqrt{7}$. Since $x \ge 1$, both $-2+\sqrt{7} \approx 0.646$ and $-2-\sqrt{7} \approx -4.646$ are rejected. $II$. Case $-3 \le x $x(x+3) - (x-1) - 2 = 0 \implies x^2+3x-x+1-2=0 \implies x^2+2x-1=0$. Using the quadratic formula, $x = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = -1 \pm \sqrt{2}$. Since $-3 \le x $III$. Case $x $x(-(x+3)) - (x-1) - 2 = 0 \implies -x^2-3x-x+1-2=0 \implies -x^2-4x-1=0 \implies x^2+4x+1=0$. Using the quadratic formula, $x = \frac{-4 \pm \sqrt{16 - 4(1)(1)}}{2} = -2 \pm \sqrt{3}$. Since $x Thus, the valid solutions are $x = -1+\sqrt{2}$, $x = -1-\sqrt{2}$, and $x = -2-\sqrt{3}$. The total number of real solutions is $3$.

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

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