# How do you solve 6= \sqrt { - 5- 3u } + 2?

May 29, 2017

u = - 7

#### Explanation:

subtract (-2) from both sides, we get

$6 - 2 = \sqrt{- 5 - 3 u} + 2 - 2$

$\Rightarrow 4 = \sqrt{- 5 - 3 u}$ [ Note: squaring both sides]

$\Rightarrow {4}^{2} = {\left[\sqrt{- 5 - 3 u}\right]}^{2}$

$\Rightarrow 16 = - 5 - 3 u$

$\Rightarrow 3 u = - 5 - 16 \mathmr{and} - 21$

$\Rightarrow u - \frac{21}{3} \mathmr{and} - 7$

May 29, 2017

$u = - 7$

#### Explanation:

$6 = \sqrt{- 5 - 3 u} + 2$

We can use some simple algebra to work this out.

$6 = \sqrt{- 5 - 3 u} + 2$

$6 - 2 = \sqrt{- 5 - 3 u}$

$4 = \sqrt{- 5 - 3 u}$

${4}^{2} = - 5 - 3 u$

$16 = - 5 - 3 u$

$16 = - 5 - 3 u$

$16 + 5 = - 3 u$

$21 = - 3 u$

u = 21 ÷ -3

color(blue)(u = -7

We can now substitute $u$ for $- 7$ to prove that we are correct.

$6 = \sqrt{- 5 - 3 u} + 2$

$6 = \sqrt{- 5 - 3 \times 7} + 2$

$6 = \sqrt{16} + 2$

$6 = 4 + 2$

$6 = 6$