# How do you simplify: Square root of 7 + square root of 7^2 + square root of 7^3 + square root of 7^4 + square root of 7^5?

Jul 24, 2015

Here's how you could simplify this expression.

#### Explanation:

$\sqrt{7} + \sqrt{7 {\text{^2) + sqrt(7""^3) + sqrt(7""^4) + sqrt(7}}^{5}}$

Now, one way in which you can simplify this expression is by using the product property of radicals, which tells you that

$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$

for $a \ge 0$ and $b \ge 0$.

This means that you can write all the numbers that are under the radicals as products of 7""^2, starting with

$\textcolor{b l u e}{7 {\text{^3 = 7}}^{2} \cdot 7}$
$\textcolor{g r e e n}{7 {\text{^4 = 7""^2 * 7}}^{2}}$
$\textcolor{\mathmr{and} a n \ge}{7 {\text{^5 = 7""^2 * 7}}^{2} \cdot 7}$

$\sqrt{7} + 7 + \textcolor{b l u e}{\sqrt{7 {\text{^2 * 7)) + color(green)(sqrt(7""^2 * 7""^2)) + color(orange)(sqrt(7""^2 * 7}}^{2} \cdot 7}}$

$\sqrt{7} + 7 + \textcolor{b l u e}{\sqrt{7 {\text{^2) * sqrt(7)) + color(green)(sqrt(7""^2) * sqrt(7""^2)) + color(orange)(sqrt(7""^2) * sqrt(7}}^{2}} \cdot \sqrt{7}}$

$\sqrt{7} + 7 + \textcolor{b l u e}{7 \sqrt{7}} + \textcolor{g r e e n}{7 \cdot 7} + \textcolor{\mathmr{and} a n \ge}{7 \cdot 7 \cdot \sqrt{7}}$

$\sqrt{7} + 7 + 7 \sqrt{7} + 49 + 49 \sqrt{7}$

Now simply add the terms that do not contain a radical and the terms that do contain a radical separately to get

$\left(7 + 49\right) + \left(\sqrt{7} + 7 \sqrt{7} + 49 \sqrt{7}\right) = 56 + 57 \sqrt{7}$

Alternatively, you can simplify this expression by grouping some of these terms together and factorizing all but the first term. This will get you

$\sqrt{7} + \left(\sqrt{7 {\text{^2) + sqrt(7""^3)) + (sqrt(7""^4) + sqrt(7}}^{5}}\right)$

$\sqrt{7} + \sqrt{7 {\text{^2) * (1 + sqrt(7)) + sqrt(7}}^{4}} \cdot \left(1 + \sqrt{7}\right)$

This is equivalent to

$\sqrt{7} + \left(1 + \sqrt{7}\right) \cdot \left(\sqrt{7 {\text{^2) + sqrt(7}}^{4}}\right)$

$\sqrt{7} + \left(1 + \sqrt{7}\right) \cdot \left(7 + 49\right)$

$\sqrt{7} + \left(1 + \sqrt{7}\right) \cdot 56$

$\sqrt{7} + 56 + 56 \cdot \sqrt{7}$

Once again, the result will be

$56 + 56 \sqrt{7} + \sqrt{7} = 56 + 57 \sqrt{7}$