# How do you solve sqrt(6a-6)=a+1 and check your solution?

Sep 30, 2017

No solutions

or using imaginary numbers
$a = 2 \pm \sqrt{3} i$

#### Explanation:

To start we would square both sides to get rid of the square root.

${\left(\sqrt{6 a - 6}\right)}^{2} = {\left(a + 1\right)}^{2}$
$6 a - 6 = {\left(a + 1\right)}^{2}$

We can now expand the brackets.

$6 a - 6 = {a}^{2} + 2 a + 1$

Move everything to one side to get the equation equal to 0.

$0 = {a}^{2} - 4 a + 7$

To solve this we will use the quadratic formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$a = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(7\right)}}{2 \left(1\right)}$
$a = \frac{4 \pm \sqrt{16 - 28}}{2}$
$a = \frac{4 \pm \sqrt{- 12}}{2}$

However we are left with the square root of a negative, so there are no solutions.

Whereas if you are using imaginary number we can continue. If you are not using imaginary numbers stop here.

We can split $\sqrt{- 12}$ into $\sqrt{- 1} \sqrt{12}$

$a = \frac{4 \pm \sqrt{- 12}}{2}$
$a = \frac{4 \pm \sqrt{- 1} \sqrt{12}}{2}$

Put $i$ in as $\sqrt{- 1}$ and simplify $\sqrt{12}$.

$a = \frac{4 \pm 2 \sqrt{3} i}{2}$
$a = 2 \pm \sqrt{3} i$