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(A) We are given the equation $x^3-3|x|+2=0$. We need to find the number of positive real solutions.Case $I$: $x > 0$Since $x > 0$,$|x| = x$. The equa

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(A) We are given the equation $x^3-3|x|+2=0$. We need to find the number of positive real solutions.Case $I$: $x > 0$Since $x > 0$,$|x| = x$. The equation becomes:$x^3 - 3x + 2 = 0$By inspection,$x = 1$ is a root. Using synthetic division or polynomial division,we can factor the expression:$(x - 1)(x^2 + x - 2) = 0$$(x - 1)(x - 1)(x + 2) = 0$$(x - 1)^2(x + 2) = 0$The roots are $x = 1$ and $x = -2$.Since we assumed $x > 0$,only $x = 1$ is a valid positive solution.Case $II$: $x Since $x $x^3 - 3(-x) + 2 = 0$$x^3 + 3x + 2 = 0$Let $f(x) = x^3 + 3x + 2$. Since $f'(x) = 3x^2 + 3 > 0$ for all $x$,the function is strictly increasing and has only one real root. Since $f(-1) = -1 - 3 + 2 = -2$ and $f(0) = 2$,the root lies in the interval $(-1, 0)$. This root is negative,so it does not satisfy the condition of being a positive real number.Thus,the only positive real number satisfying the equation is $x = 1$.Therefore,there is only $1$ positive real solution.

How many positive real numbers $x$ satisfy the equation $x^3-3|x|+2=0$?

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