How do you solve #x+3=sqrt(2x^2-7)# and identify any restrictions?

Question

2132 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-03T21:45:04

Range restriction: #x>=+-sqrt(14)/2 ~~1.87#

Solutions (roots): #x=-2 and x=8#

#x+3=sqrt(2x^2-7)#

First let's look at the restrictions, the value within the radical cannot be negative or the solution is imaginary:

#2x^2-7>=0#

#x^2>=7/2#

#x>=+-sqrt(7/2)#

#x>=+-sqrt(14)/2 ~~1.87#

Now solve:

#x+3=sqrt(2x^2-7)#

#(x+3)^2=(sqrt(2x^2-7))^2#

#x^2 +6x+9=2x^2-7#

#0=x^2-6x-16#

#(x+2)(x-8)=0#

#x=-2 and x=8#