How do you solve $w^{2} - 2 \leq 0$ $w^2-2<=0$ using a sign chart?

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Answer 1 · 2017-09-26T12:43:38

Solution: $- \sqrt{2} \leq w \leq \sqrt{2} or [- \sqrt{2}, \sqrt{2}]$ $-sqrt2 <= w <= sqrt2 or [-sqrt2,sqrt2]$

$w^{2} - 2 \leq 0 or (w + \sqrt{2}) (w - \sqrt{2}) \leq 0$ $w^2-2 <=0 or (w+sqrt2)(w-sqrt2) <= 0$ . Critical points

are $w = \sqrt{2} and w = - \sqrt{2}$ $w= sqrt2 and w = -sqrt2$ at $w = \pm \sqrt{2}; w^{2} - 2 = 0$ $w = +- sqrt2 ; w^2-2=0$

Sign chart:

When $w < - \sqrt{2}$ $w < -sqrt2$ sign of $(w + \sqrt{2}) (w - \sqrt{2})$ $(w+sqrt2)(w-sqrt2)$ is $(-) \cdot (-) = +; > 0$ $(-)*(-) =+ ; >0$

When $- \sqrt{2} < w < \sqrt{2}$ $-sqrt2 < w < sqrt2$ sign of $(w + \sqrt{2}) (w - \sqrt{2})$ $(w+sqrt2)(w-sqrt2)$

is $(+) \cdot (-) = -; < 0$ $(+)*(-) =- ; < 0$

When $w > \sqrt{2}$ $w > sqrt2$ sign of $(w + \sqrt{2}) (w - \sqrt{2})$ $(w+sqrt2)(w-sqrt2)$ is $(+) \cdot (+) = +; > 0$ $(+)*(+) =+ ; >0$

Solution: $- \sqrt{2} \leq w \leq \sqrt{2} or [- \sqrt{2}, \sqrt{2}]$ $-sqrt2 <= w <= sqrt2 or [-sqrt2,sqrt2]$

graph{x^2-2 [-12.66, 12.65, -6.33, 6.33]} [Ans]

How do you solve w^2-2<=0w2−2≤0w^2-2<=0 using a sign chart?