How do you solve #x=sqrt(2x^2-8)+2# and identify any restrictions?
Our restriction is that
Square both sides:
Expand the left side:
Combine like terms on the left side:
Move everything on the left side to the right side:
Combine like terms on the right side:
Solve for x:
Let's check the solutions:
And! We have restrictions:
- we can't allow the term under the square root to be less than 0. What values of
Which means that
- 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