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

Jun 3, 2018

Range restriction: $x \ge \pm \frac{\sqrt{14}}{2} \approx 1.87$

Solutions (roots): $x = - 2 \mathmr{and} x = 8$

Explanation:

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

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

$2 {x}^{2} - 7 \ge 0$

${x}^{2} \ge \frac{7}{2}$

$x \ge \pm \sqrt{\frac{7}{2}}$

$x \ge \pm \frac{\sqrt{14}}{2} \approx 1.87$

Now solve:

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

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

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

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

$\left(x + 2\right) \left(x - 8\right) = 0$

$x = - 2 \mathmr{and} x = 8$