How do you simplify $\sqrt{a^{2}}$ $sqrt(a^2)$ ?

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How do you simplify $\sqrt{a^{2}}$ $sqrt(a^2)$ ?

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Answer 1 · 2018-04-15T13:33:25

$a$ $a$
Refer to the explanation.

Explanation:

$\sqrt{a^{2}} \Rightarrow a^{\frac{2}{2}} \Rightarrow a$ $sqrt(a^2) rArr a^(2/2) rArr a$
law of indices : $\sqrt[n]{a^{m}} \Rightarrow a^{\frac{m}{n}}$ $root(n)(a^m) rArr a^(m/n)$

Hope this helps :)

Answer 2 · 2018-04-15T14:31:35

See below.

Explanation:

To be more accurate, $\sqrt{a^{2}} = | a |$ $sqrt(a^2) = abs a$ ...

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

Case 1: $a > 0$ $a>0$

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

Case 2: $a < 0$ $a<0$

Let $a = - 3$ $a = -3$ . Then $\sqrt{a^{2}} = \sqrt{{(- 3)}^{2}} = \sqrt{9} = 3 \neq a$ $sqrt (a^2) = sqrt ((-3)^2) = sqrt 9 = 3 != a$ . In this case, $\sqrt{a^{2}} \neq a$ $sqrt (a^2) != a$ . However, it does equal $| a |$ $abs a$ because $| - 3 | = 3$ $abs (-3) = 3$ .

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

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