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

2 Answers

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

Answer:

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).