# How do you solve the equation and identify any extraneous solutions for sqrt(x^2 - 5) = 4?

Jun 11, 2015

An extraneous solution would be if the term under the root were $< 0$ which means that ${x}^{2} \ge 5 \to x \ge \sqrt{5} \mathmr{and} x \le - \sqrt{5}$
${x}^{2} - 5 = {4}^{2} \to {x}^{2} = 21 \to x = \sqrt{21} \mathmr{and} x = - \sqrt{21}$