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Here is the question!
@hero wbu?
A couple rules to keep in mind before approaching this: \(\sqrt{x^2} = x\) (because square and square root are inverses of each other) \(\sqrt{ab} = \sqrt{a} \times \sqrt{b}\) (related to rules of exponents) Now given: \(x^2 = 160\) Square root both sides: \(\sqrt{x^2} = \sqrt{160}\) \(x = \sqrt{(16)(10)}\) \(x = \sqrt{16} \times \sqrt{10}\)
You can finish simplifying the result from here.
and when you take square root on both sides, don't forget to consider the negative root too :) \(\sqrt{x^2} =\pm \sqrt{160}\)
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Thanks guys idk who to give a medal to though? :/
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