How do you solve $x^{2} - 4 \leq 0$ $x^2-4<=0$ ?

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How do you solve $x^{2} - 4 \leq 0$ $x^2-4<=0$ ?

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Answer 1 · 2016-08-30T14:47:33

$- 2 \leq x \leq 2$ $-2 ≤ x ≤2$

Explanation:

Write as a quadratic equation and solve, and then select test points.

$x^{2} - 4 = 0$ $x^2 - 4 =0$

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

$x = - 2 and 2$ $x = -2 and 2$

Usually, we choose one point inside the parabola and another outside

The x-intercepts of the parabola are $x = \pm 2$ $x = +-2$ , so let the test point on the exterior be $x = - 4$ $x = -4$ and the point on the interior be $x = 1$ $x = 1$ .

Test point 1: $x = - 4$ $x = -4$

$- 4^{2} - 4 \leq^{?} 0$ $-4^2 - 4 <=^?0$

$12 cancel(<=) 0$

Hence, by deduction, the interval of solution is $-2 ≤ x ≤ 2$ . Or, in other words, the inside of the parabola is shaded.

Hopefully this helps!

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How do you solve $x^{2} - 4 \leq 0$ $x^2-4<=0$ ?

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