Question

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

Answer 1

`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`