# How do you solve \sqrt { x ^ { 2} - 2x + 4} = \sqrt { x ^ { 2} + 5x - 12}?

Jul 21, 2017

$x = 2 \frac{2}{7}$

#### Explanation:

Step 1) Square both sides

$\sqrt{{x}^{2} - 2 x + 4} = \sqrt{{x}^{2} + 5 x - 12}$

${x}^{2} - 2 x + 4 = {x}^{2} + 5 x - 12$

Step 2) Put the variables on the left side and the real numbers on the right

$- 2 x + 4 = + 5 x - 12$

$- 2 x - 5 x = - 12 - 4$

$- 7 x = - 16$

Step 3) Solve

$x = \frac{- 16}{-} 7$

$x = 2 \frac{2}{7}$

Jul 21, 2017

$x = \frac{16}{7} = 2 \frac{2}{7}$

#### Explanation:

$\sqrt{{x}^{2} - 2 x + 4} = \sqrt{{x}^{2} + 5 x - 12}$

Squaring the two sides

${x}^{2} - 2 x + 4 = {x}^{2} + 5 x - 12$

i.e. $5 x + 2 x = 4 + 12$

or $7 x = 16$

i.e. $x = \frac{16}{7} = 2 \frac{2}{7}$

Jul 21, 2017

$x = \frac{16}{7}$

#### Explanation:

As there s a square root on each sidez we just square each side to remove the square root:
${\left(\sqrt{{x}^{2} - 2 x + 4}\right)}^{2} = {\left(\sqrt{{x}^{2} + 5 x - 12}\right)}^{2}$

${x}^{2} - 2 x + 4 = {x}^{2} + 5 x - 12$

As each side has one positive ${x}^{2}$, we can just remove them by subtracting ${x}^{2}$:

${x}^{2} - 2 x + 4 - \textcolor{red}{{x}^{2}} = {x}^{2} + 5 x - 12 - \textcolor{red}{{x}^{2}}$

$- 2 x + 4 = 5 x - 12$

Now, we can put the $x$s onto the RHS by adding $2 x$:

$- 2 x + 4 + \textcolor{red}{2 x} = 5 x - 12 + \textcolor{red}{2 x}$

$4 = 7 x - 12$

Now we just need the integers on the LHS by adding $12$:

$4 + \textcolor{red}{12} = 7 x - 12 + \textcolor{red}{12}$

$16 = 7 x$

To get $x$ on its own, we just divide both sides by 7:

$\frac{16}{\textcolor{red}{7}} = \frac{\cancel{7} x}{\cancel{\textcolor{red}{7}}}$

$x = \frac{16}{7}$