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

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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:

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

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

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

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

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

$x=-2 and x=8$

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