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

1 Answer
Astralboy · Eddie · MathFact-orials.blogspot.com
Apr 2, 2017

#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#

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