Answer: [tex]\Big\{\pm 4\sqrt{5} \Big\}[/tex]
This is the shorthand way of writing [tex]\Big\{4\sqrt{5}, -4\sqrt{5} \Big\}[/tex]
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Work Shown:
[tex]x^2 - 80 = 0\\\\x^2 = 80\\\\x = \pm\sqrt{80}\\\\x = \pm\sqrt{16*5}\\\\x = \pm\sqrt{16}*\sqrt{5}\\\\x = \pm4\sqrt{5}\\\\x = 4\sqrt{5} \ \text{ or } \ x = -4\sqrt{5}\\\\[/tex]
The plus or minus is needed because squaring a negative leads to a positive. As another example, x^2 = 25 has x = 5 and x = -5 as the two solutions. Note that x^2 = (-5)^2 = (-5)*(-5) = 25.
Also, note that the 80 was broken up into 16*5. This was done to simplify the square root. We pull out the largest factor that's a perfect square.
So that's how we get to the solution set [tex]\Big\{\pm 4\sqrt{5} \Big\}[/tex]. The curly braces tell the reader that they are dealing with a set.
