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

1 Answer
Mar 20, 2017

Answer:

#x=-2#

Explanation:

Square both sides:

#sqrt(3x+10)^2=(sqrt(x+11)-1)^2#

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

#3x+10#

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

#(sqrt(x+11))^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 - 2sqrt(x+11)#

Now write both sides together:

#3x+10=x+12 - 2sqrt(x+11)#

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

#(2x-2)/-2=sqrt(x+11)#

Let's simplify the left side:

#(2x-2)/-2=-x+1#

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

#(-x+1)^2=(sqrt(x+11))^2#

The square root cancels with the square:

#(-x+1)^2=x+11#

Distribute on the left side:

#x^2-x-x+1=x+11#

Simplify the left side:

#x^2-2x+1=x+11#

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

#x^2-3x-10=0#

Factor:

#(x-5)(x+2)=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#