# How do you solve x+5=sqrt(17+x)??

Jun 22, 2018

$x = - 1$

#### Explanation:

$x + 5 = \sqrt{17 + x}$

Square both sides
${x}^{2} + 10 x + 25 = 17 + x$

Subtract to make one side 0
${x}^{2} + 9 x + 8 = 0$

Factor
$\left(x + 8\right) \left(x + 1\right) = 0$

Solve for x:
$x + 8 = 0 \mathmr{and} x + 1 = 0$
$x = - 8 \mathmr{and} x = - 1$

Since we squared both sides, we must check for extraneous solutions:
$- 1 + 5 = \sqrt{17 - 1}$
$4 = 4$, $x = - 1$

$- 8 + 5 = \sqrt{17 - 8}$
$- 3 = \sqrt{9}$
By convention, the $\sqrt{}$ symbol always refers to the positive square root, so:
$- 3 \ne 3$, $x \ne - 8$

Therefore, $x = - 1$

Jun 22, 2018

$x = - 1$

#### Explanation:

$x + 5 = \sqrt{17 + x}$

First, square both sides:
${\left(x + 5\right)}^{\textcolor{b l u e}{2}} = {\left(\sqrt{17 + x}\right)}^{\textcolor{b l u e}{2}}$

${\left(x + 5\right)}^{2} = 17 + x$

We know that any binomial in the form of:
${\left(x + y\right)}^{2} = {x}^{2} + 2 x y + {y}^{2}$, so the left hand side becomes:
${x}^{2} + 10 x + 25$

Put that back into the equation:
${x}^{2} + 10 x + 25 = 17 + x$

Subtract $\textcolor{b l u e}{17}$ and $\textcolor{b l u e}{x}$ from both sides:
${x}^{2} + 9 x + 8 = 0$

This is in standard form, or $\textcolor{red}{a} {x}^{2} + \textcolor{g r e e n}{b} x + \textcolor{m a \ge n t a}{c}$. In this form, we can factor it with two numbers that:

1. Multiply up to $\textcolor{red}{a} \cdot \textcolor{m a \ge n t a}{c} = \textcolor{red}{1} \cdot \textcolor{m a \ge n t a}{8} = 8$
2. Add up to $\textcolor{g r e e n}{b}$, or $\textcolor{g r e e n}{9}$

Those two numbers are $\textcolor{b l u e}{8}$ and $\textcolor{b l u e}{1}$, as:

1. $1 \cdot 8 = 8$
2. $1 + 8 = 9$

Therefore, the factored form is:
$\left(x + 8\right) \left(x + 1\right) = 0$

Since when the expressions multiply to equal zero, that means each of the expressions equal zero. So we can set up two equations:
$x + 8 = 0$ and $x + 1 = 0$

Subtract $\textcolor{b l u e}{8}$ in the first equation, subtract $\textcolor{b l u e}{1}$ in the second equation:
$x = - 8$ and $x = - 1$

$- - - - - - - - - - - - - - - - - - - -$

Now we have to check if both solutions are really solutions by plugging them into the original formula. Let's check $\textcolor{b l u e}{- 8}$ first:

$x + 5 = \sqrt{17 + x}$

$- 8 + 5 = \sqrt{17 - 8}$

$- 3 = \sqrt{9}$

$- 3 \ne 3$

No, this is not true, so $\textcolor{b l u e}{- 8}$ is NOT a solution! Now check $\textcolor{b l u e}{- 1}$:
$- 1 + 5 = \sqrt{17 - 1}$

$4 = \sqrt{16}$

$4 = 4$

Yes, this is true, so $\textcolor{b l u e}{- 1}$ is the ONLY solution!

Hope this helps!