# How do you solve X + sqrt( X + 5) = 7?

We can write it as follows

x+sqrt(x+5)=7=>sqrt(x+5)=7-x=>(sqrt(x+5))^2=(7-x)^2=> x+5=49-14x+x^2=> x^2-15x+44=0=> (x-11)*(x-4)=0=> x=4 or x=11

The only acceptable solution is x=4

Sep 9, 2015

Rearrange, square both sides and solve the resulting quadratic to find two possible solutions, one of which is spurious and the other a solution of the original equation: $X = 4$

#### Explanation:

First subtract $X$ from both sides to get:

$\sqrt{X + 5} = 7 - X$

Then square both sides to get:

$X + 5 = {\left(7 - X\right)}^{2} = 49 - 14 X + {X}^{2}$

Note that squaring both sides may have introduced a spurious solution, so we need to check later.

Next subtract $X + 5$ from both sides to get:

${X}^{2} - 15 X + 44 = 0$

Now $44 = 4 \cdot 11$ and $15 = 4 + 11$, so

${X}^{2} - 15 X + 44 = \left(X - 4\right) \left(X - 11\right)$

which is zero when $X = 4$ or $X = 11$.

Substituting these into the left hand side of the original equation we find:

$4 + \sqrt{4 + 5} = 4 + \sqrt{9} = 4 + 3 = 7$ so $X = 4$ is a solution.

$11 + \sqrt{11 + 5} = 11 + \sqrt{16} = 11 + 4 = 15 \ne 7$ so $X = 11$ is spurious.

(Note: $11 - \sqrt{16} = 7$ so obviously this spurious 'solution' came from squaring the square root).