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

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Answer 1 · 2018-06-03T21:45:04

Range restriction: $x \geq \pm \frac{\sqrt{14}}{2} \approx 1.87$ $x>=+-sqrt(14)/2 ~~1.87$

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

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

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

$2 x^{2} - 7 \geq 0$ $2x^2-7>=0$

$x^{2} \geq \frac{7}{2}$ $x^2>=7/2$

$x \geq \pm \sqrt{\frac{7}{2}}$ $x>=+-sqrt(7/2)$

$x \geq \pm \frac{\sqrt{14}}{2} \approx 1.87$ $x>=+-sqrt(14)/2 ~~1.87$

Now solve:

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

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

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

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

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

$x = - 2 and x = 8$ $x=-2 and x=8$

How do you solve x+3=√2x2−7x+3=sqrt(2x^2-7) and identify any restrictions?