The solution set of x^{2} leq 4 is...... | vedclass

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(A) To solve the inequality $x^{2} \leq 4$,we can rewrite it as $x^{2} - 4 \leq 0$.This is a difference of squares,which factors as $(x - 2)(x + 2) \l

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$[-2, 2]$

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(A) To solve the inequality $x^{2} \leq 4$,we can rewrite it as $x^{2} - 4 \leq 0$.This is a difference of squares,which factors as $(x - 2)(x + 2) \leq 0$.The roots of the corresponding equation $(x - 2)(x + 2) = 0$ are $x = 2$ and $x = -2$.These roots divide the number line into three intervals: $(-\infty, -2)$,$(-2, 2)$,and $(2, \infty)$.Testing a value in the interval $(-2, 2)$,such as $x = 0$: $(0 - 2)(0 + 2) = -4$,which is $\leq 0$.Since the inequality includes the equality sign,the endpoints are included.Thus,the solution set is the closed interval $[-2, 2]$.

The solution set of $x^{2} \leq 4$ is......

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