How do you simplify sqrt(a^2)?

Apr 15, 2018

$a$
Refer to the explanation.

Explanation:

$\sqrt{{a}^{2}} \Rightarrow {a}^{\frac{2}{2}} \Rightarrow a$
law of indices : $\sqrt[n]{{a}^{m}} \Rightarrow {a}^{\frac{m}{n}}$

Hope this helps :)

Apr 15, 2018

See below.

Explanation:

To be more accurate, $\sqrt{{a}^{2}} = \left\mid a \right\mid$...

Let's consider two cases: $a > 0$ and $a < 0$.

Case 1: $a > 0$

Let $a = 3$. Then $\sqrt{{a}^{2}} = \sqrt{{3}^{2}} = \sqrt{9} = 3 = a$.
In this case, $\sqrt{{a}^{2}} = a$.

Case 2: $a < 0$

Let $a = - 3$. Then $\sqrt{{a}^{2}} = \sqrt{{\left(- 3\right)}^{2}} = \sqrt{9} = 3 \ne a$. In this case, $\sqrt{{a}^{2}} \ne a$. However, it does equal $\left\mid a \right\mid$ because $\left\mid - 3 \right\mid = 3$.

Whether $a > 0$ or $a < 0$, $\sqrt{{a}^{2}} > 0$; it will always be positive. We account for this with the absolute value sign: $\sqrt{{a}^{2}} = \left\mid a \right\mid$.