# How do you solve sqrt( 2x+1)=x-3?

Aug 5, 2015

#### Answer:

This equation has 2 solutions: ${x}_{1} = 4 - 2 \sqrt{2}$ and ${x}_{2} = 4 + 2 \sqrt{2}$

#### Explanation:

In the equation there is a square root, so first you have to calculate the domain.

In this case the experssion under root sign is $2 x + 1$ so you have the domain: $2 x + 1 \ge 0$ or $x \ge - \frac{1}{2}$

Now we can calculate the solutions.
We start with raising both sides to the second power

$2 x + 1 = {\left(x - 3\right)}^{2}$

$2 x + 1 = {x}^{2} - 6 x + 9$

$- {x}^{2} + 8 x - 8 = 0$

Now we can solve the quadratic equation:

$\Delta = {8}^{2} - 4 \cdot \left(- 1\right) \cdot \left(- 8\right)$

$\Delta = 64 - 32 = 32$

$\sqrt{\Delta} = \sqrt{32} = 4 \sqrt{2}$

${x}_{1} = \frac{- 8 - 4 \sqrt{2}}{2 \cdot \left(- 1\right)}$

${x}_{1} = 4 + 2 \sqrt{2}$

${x}_{2} = \frac{- 8 + 4 \sqrt{2}}{2 \cdot \left(- 1\right)}$

${x}_{2} = 4 - 2 \sqrt{2}$

Now we have to check if any of calculated solutions are in the domain.

${x}_{1} > 4$ so it surely is greater than $- \frac{1}{2}$

${x}_{2} = 4 - 2 \sqrt{2} \approx 4 - 2 \cdot 1 , 41 \approx 4 - 2 , 82 \approx 1 , 18 \ge - \frac{1}{2}$

We checked that both solutions are in the domain, so we can write the answer:

This equation has 2 solutions: ${x}_{1} = 4 - 2 \sqrt{2}$ and ${x}_{2} = 4 + 2 \sqrt{2}$