How do you solve the equation and identify any extraneous solutions for sqrt(x^2 + 2) = x + 4?

Jun 19, 2015

I found $x = - \frac{7}{4}$

Explanation:

I would square both sides to get:
${x}^{2} + 2 = {\left(x + 4\right)}^{2}$
$\cancel{{x}^{2}} + 2 = \cancel{{x}^{2}} + 8 x + 16$
$8 x = - 14$
$x = - \frac{14}{8}$
$x = - \frac{7}{4}$

Jun 19, 2015

Square both sides to get:

${x}^{2} + 2 = {x}^{2} + 8 x + 16$

Hence $x = - \frac{7}{4}$

Explanation:

Try squaring both sides to get:

${x}^{2} + 2 = {\left(x + 4\right)}^{2} = {x}^{2} + 8 x + 16$

Subtract ${x}^{2} + 16$ from both sides to get:

$8 x = - 14$

Divide both sides by $8$ to get:

$x = - \frac{14}{8} = - \frac{7}{4}$

Check:

$\sqrt{{x}^{2} + 2} = \sqrt{\frac{49}{16} + \frac{32}{16}} = \sqrt{\frac{81}{16}} = \frac{9}{4}$

$x + 4 = - \frac{7}{4} + 4 = - \frac{7}{4} + \frac{16}{4} = \frac{9}{4}$

graph{(y - sqrt(x^2+2))*(y - x - 4) = 0 [-9.42, 10.58, -1.8, 8.2]}