# How do you write sqrt(24a^10b^6) in simplest form?

Mar 24, 2015

You can write it as:
$\sqrt{6 \cdot 4 \cdot {a}^{10} {b}^{6}} = 2 {a}^{5} {b}^{3} \sqrt{6}$

Where you used the fact that $\sqrt{x} = {x}^{\frac{1}{2}}$

and so:
$\sqrt{6 \cdot 4 \cdot {a}^{10} {b}^{6}} = {\left({2}^{2} \cdot 6 \cdot {a}^{10} {b}^{6}\right)}^{\frac{1}{2}} =$
$= {2}^{2 \cdot \frac{1}{2}} {6}^{\frac{1}{2}} {a}^{10 \cdot \frac{1}{2}} {b}^{6 \cdot \frac{1}{2}} = 2 {a}^{5} {b}^{3} \sqrt{6}$

Mar 24, 2015

If we may assume that $a$ and $b$ are non-negative, then we write $\sqrt{24 {a}^{10} {b}^{6}} = 2 {a}^{5} {b}^{3} \sqrt{6}$

$\sqrt{24 {a}^{10} {b}^{6}} = \sqrt{4 \cdot 6 \cdot {\left({a}^{5}\right)}^{2} {\left({b}^{3}\right)}^{2}} = 2 \sqrt{6} {a}^{5} {a}^{3} = 2 {a}^{5} {b}^{3} \sqrt{6}$

The key is the repeated use of: $\sqrt{{n}^{2}} = n$
(for non-negative $n$).

So $\sqrt{4} = \sqrt{{2}^{2}} = 2$, and $\sqrt{{\left({a}^{5}\right)}^{2}} = {a}^{5}$ and $\sqrt{{\left({b}^{3}\right)}^{2}} = {b}^{3}$.

(If we may NOT assume $a$ and $b$ are non-negative, the we need absolute values, because then we would need $\sqrt{{a}^{2}} = \left\mid a \right\mid$
$\sqrt{24 {a}^{10} {b}^{6}} = 2 \left\mid {a}^{5} {b}^{3} \right\mid \sqrt{6}$.)

Mar 24, 2015

Extract the largest square from each of the terms within the square root:
$24 = {2}^{2} \cdot 6$
${a}^{10} = {\left({a}^{5}\right)}^{2} \cdot \left(1\right)$
${b}^{6} = {\left({b}^{3}\right)}^{2} \cdot \left(1\right)$

So
$\sqrt{24 {a}^{10} {b}^{6}}$

$= \sqrt{\left({\textcolor{red}{2}}^{2}\right) \left(6\right) {\textcolor{red}{\left({a}^{5}\right)}}^{2} {\textcolor{red}{\left({b}^{3}\right)}}^{2}}$

$= \textcolor{red}{2 {a}^{5} {b}^{3}} \sqrt{6}$