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

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

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Answer 1 · 2015-06-11T12:12:18

An extraneous solution would be if the term under the root were $< 0$ $<0$ which means that $x^{2} \geq 5 \to x \geq \sqrt{5} or x \leq - \sqrt{5}$ $x^2>=5->x>=sqrt5orx<=-sqrt5$

Explanation:

Now we can square both sides:
$x^{2} - 5 = 4^{2} \to x^{2} = 21 \to x = \sqrt{21} or x = - \sqrt{21}$ $x^2-5=4^2->x^2=21->x=sqrt21orx=-sqrt21$

These both meet the conditions at the top.

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

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How do you solve the equation and identify any extraneous solutions for sqrt(x^2 - 5) = 4√x2−5=4sqrt(x^2 - 5) = 4?

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