# How do you evaluate square root of a^(11) + sqrt(a^5)?

Jul 6, 2017

${a}^{\frac{5}{2}} \left({a}^{\frac{17}{2}} + 1\right)$ This may also be written as: $\sqrt{{a}^{5}} \left(\sqrt{{a}^{17}} + 1\right)$

The use of the word 'evaluate' in my book states; 'give value to'. This is not possible unless you know the value of $a$

#### Explanation:

$\textcolor{b l u e}{\text{Some facts about indices and roots }}$

Consider the example $\sqrt{b}$ this may also be written as ${b}^{\frac{1}{2}}$

$\sqrt[3]{b} = {b}^{\frac{1}{3}}$
$\sqrt[4]{b} = {b}^{\frac{1}{4}}$
$\sqrt[5]{b} = {b}^{\frac{1}{5}}$
So the root is the denominator of the 'power'

$\sqrt{{b}^{2}} = {b}^{\frac{2}{2}} = {b}^{1} = b$
$\sqrt{{b}^{3}} = {b}^{\frac{3}{2}}$
$\sqrt{{b}^{4}} = {b}^{\frac{4}{2}} = {b}^{2}$
and so on

Consider the example ${3}^{3} \times 3 \text{ " =" " 3xx3xx3xx3" "=" } {3}^{4}$

3 may, if you so wish, be written as ${3}^{1}$

So ${3}^{2} \times 3 \text{ "=3^2xx3^1" "=" "3^(3+1)" "=" } {3}^{4}$

So ${b}^{5} \times {b}^{7} = {b}^{5 + 7} = {b}^{12}$

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$\textcolor{b l u e}{\text{Answering the question}}$

Given:$\text{ } {a}^{11} + \sqrt{{a}^{5}}$

Write as:

$\left[{a}^{11}\right] + \left[\sqrt{{x}^{5}} \times 1\right]$

But ${a}^{11}$ is the same as ${a}^{\frac{22}{2}} \text{ "=" "a^(5/2+17/2)" "=" } {a}^{\frac{5}{2}} \times {a}^{\frac{17}{2}}$

Also $\sqrt{{a}^{5}}$ is the same as ${a}^{\frac{5}{2}}$

Putting it all together we have

$\left[{a}^{\frac{5}{2}} \times {a}^{\frac{17}{2}}\right] + \left[{a}^{\frac{5}{2}} \times 1\right]$

Factor out the ${a}^{\frac{5}{2}}$ giving:

${a}^{\frac{5}{2}} \left({a}^{\frac{17}{2}} + 1\right)$

This may also be written as:

$\sqrt{{a}^{5}} \left(\sqrt{{a}^{17}} + 1\right)$

NOT CONVINCED THIS HELPS