How do you solve #x^2-4<=0#?

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Answer 1 · 2016-08-30T14:47:33

#-2 ≤ x ≤2#

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

#x^2 - 4 =0 #

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

#x = -2 and 2#

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

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

Test point 1: #x = -4#

#-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!