Question

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

Answer 1

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

Answer 2

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 = (7-X)^2 = 49-14X+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^2-15X+44 = 0`

Now `44 = 4 * 11` and `15 = 4 + 11`, so

`X^2-15X+44 = (X-4)(X-11)`

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 != 7` so `X=11` is spurious.

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