# How do you simplify sqrt(-15)( sqrt(-4 -sqrt7))?

Apr 16, 2017

$\sqrt{- 15} \sqrt{- 4 - \sqrt{7}} = - \frac{\sqrt{30}}{2} - \frac{\sqrt{210}}{2}$

#### Explanation:

By definition and convention, the principal square root of a negative real number $x$ is given by:

$\sqrt{x} = i \sqrt{- x}$

where $i$ is the imaginary unit, satisfying ${i}^{2} = - 1$.

If $a , b \ge 0$ then:

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

As a first step in attempting to simplify the given expression, let's try to find a simpler expression for:

$\sqrt{4 + \sqrt{7}}$

This may be expressible in the form $a + b \sqrt{7}$. Let's see what we get when we square $a + b \sqrt{7}$...

${\left(a + b \sqrt{7}\right)}^{2} = \left({a}^{2} + 7 {b}^{2}\right) + 2 a b \sqrt{7}$

So we would like ${a}^{2} + 7 {b}^{2} = 4$ and $2 a b = 1$

Notice that if $a = b = 1$ then we would get ${a}^{2} + 7 {b}^{2} = 8$ and $2 a b = 2$, both twice the wanted values.

So we can use $a = b = \frac{\sqrt{2}}{2}$ to find:

$\sqrt{4 + \sqrt{7}} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} \sqrt{7}}{2} = \frac{\sqrt{2}}{2} + \frac{\sqrt{14}}{2}$

From our previous remarks:

$\sqrt{- 4 - \sqrt{7}} = i \sqrt{4 + \sqrt{7}} = i \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{14}}{2}\right)$

Then:

$\sqrt{- 15} \sqrt{- 4 - \sqrt{7}} = i \sqrt{15} \cdot i \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{14}}{2}\right)$

$\textcolor{w h i t e}{\sqrt{- 15} \sqrt{- 4 - \sqrt{7}}} = - \sqrt{15} \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{14}}{2}\right)$

$\textcolor{w h i t e}{\sqrt{- 15} \sqrt{- 4 - \sqrt{7}}} = - \frac{\sqrt{30}}{2} - \frac{\sqrt{210}}{2}$