How do you solve $\sqrt{4 n^{2}} = \frac{1}{3}$ $sqrt(4n^2) = 1/3$ ?

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Answer 1 · 2015-10-07T15:52:22

$n = \frac{1}{6}$ $n = 1/(6)$

Square both sides: $4 n^{2} = \frac{1}{9}$ $4n^2 = 1/9$

Divide throughout by 4: $n^{2} = \frac{1}{36}$ $n^2 = 1/(36)$

Square root both sides: $n = \pm \sqrt{\frac{1}{36}}$ $n = +- sqrt(1/(36))$

Solution: $n = \pm \frac{1}{6}$ $n = +- 1/6$

Comment: $(- n) (- n) = (+ n) (+ n) = n^{2}$ $(-n)(-n) = (+n)(+n) = n^2$ so you have to have both a positive and negative solution.

How do you solve sqrt(4n^2) = 1/3√4n2=13sqrt(4n^2) = 1/3?