Question

How do you solve `x=sqrt(2x^2-8)+2` and identify any restrictions?

Answer 1

`x=2`

Our restriction is that `x>=2`.

Explanation:

Subtract `2` on both sides:

`x-2=sqrt(2x^2-8)`

Square both sides:

`(x-2)^2=2x^2-8`

Expand the left side:

`x^2-2x-2x+4=2x^2-8`

Combine like terms on the left side:

`x^2-4x+4=2x^2-8`

Move everything on the left side to the right side:

`0=2x^2-8-(x^2-4x+4)`

Combine like terms on the right side:

`x^2 color(red)(+)4x-12=0`

Factor:

`(xcolor(red)(+)6)(x color(red)(-) 2)=0`

Solve for x:

`x=2, -6`

Let's check the solutions:

`x=sqrt(2x^2-8)+2`

`2=sqrt(2(2)^2-8)+2`

`2=sqrt(2(4)-8)+2`

`2=sqrt(8-8)+2`

`2=sqrt(0)+2`

`2=2 color(white)(00)color(green)sqrt`

`-6=sqrt(2(-6)^2-8)+2`

`-6=sqrt(2(36)-8)+2`

`-6=sqrt(72-8)+2`

`-6=sqrt(64)+2`

`-6=8+2`

`-6=10color(white)(00)color(red)(X)`

And! We have restrictions:

• we can't allow the term under the square root to be less than 0. What values of `x` are allowed?

`2x^2-8>=0`

`2x^2>=8`

`x^2>=4`

`x>=2, <=-2`

Which means that `-2 < x < 2` is disallowed.

• Also, from `x-2=sqrt(2x^2-8)`, we can see that `x-2>=0=>x>=2`. It's this restriction that eliminates `x=-6` from the solutions.

Therefore the restriction is `x>=2`