# How do you solve sqrt(3x+10)=sqrt(x+11)-1?

Mar 20, 2017

$x = - 2$

#### Explanation:

Square both sides:

${\sqrt{3 x + 10}}^{2} = {\left(\sqrt{x + 11} - 1\right)}^{2}$

The square of a square root is simply what's on the inside. So the left side is simply:

$3 x + 10$

However, the right side, you have to distribute. Use FOIL:

${\left(\sqrt{x + 11}\right)}^{2} - \sqrt{x + 11} - \sqrt{x + 11} + 1$

Again, the square root cancels with the squared:

$x + 11 - \sqrt{x + 11} - \sqrt{x + 11} + 1$

Combine like terms:

$x + 12 - 2 \sqrt{x + 11}$

Now write both sides together:

$3 x + 10 = x + 12 - 2 \sqrt{x + 11}$

Isolate the square root by subtracting $x + 12$ and then dividing by $- 2$:

$\frac{2 x - 2}{-} 2 = \sqrt{x + 11}$

Let's simplify the left side:

$\frac{2 x - 2}{-} 2 = - x + 1$

Now we have another square root so we have to square both sides again:

${\left(- x + 1\right)}^{2} = {\left(\sqrt{x + 11}\right)}^{2}$

The square root cancels with the square:

${\left(- x + 1\right)}^{2} = x + 11$

Distribute on the left side:

${x}^{2} - x - x + 1 = x + 11$

Simplify the left side:

${x}^{2} - 2 x + 1 = x + 11$

Move everything to the left by subtracing $x + 11$

${x}^{2} - 3 x - 10 = 0$

Factor:

$\left(x - 5\right) \left(x + 2\right) = 0$

Set $x - 5$ and $x + 2$ equal to 0 and solve for x:

$x = 5 , - 2$

Plug back in to check:

$\sqrt{25} = \sqrt{16} - 1$

This doesn't work so 5 is an extraneous variable. Now the other one:

$\sqrt{4} = \sqrt{9} - 1$

That works! So we only have

$x = - 2$