The number of real solutions of the equation | vedclass

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(D) Given equation: $|x|^2 - 3|x| + 2 = 0$.Let $|x| = t$. Since $|x| \ge 0$,we have $t \ge 0$.The equation becomes $t^2 - 3t + 2 = 0$.Factoring the qu

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$4$

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(D) Given equation: $|x|^2 - 3|x| + 2 = 0$.Let $|x| = t$. Since $|x| \ge 0$,we have $t \ge 0$.The equation becomes $t^2 - 3t + 2 = 0$.Factoring the quadratic equation: $(t - 1)(t - 2) = 0$.This gives $t = 1$ or $t = 2$.Case $I$: $|x| = 1 \Rightarrow x = 1, -1$.Case $II$: $|x| = 2 \Rightarrow x = 2, -2$.Thus,the real solutions are $x \in \{1, -1, 2, -2\}$.The total number of real solutions is $4$.

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

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