Algebra - Absolute Value Equations

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Section 2.14 : Absolute Value Equations

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7. Find all the real valued solutions to the equation.

\[\left| {{x^2} + 4} \right| = 1\]

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To this point we’ve only worked problems that have linear terms in the absolute value bars. In this case we have a quadratic in the absolute value bars. That doesn’t change how the problem works however. We work this exactly like the previous problems.

Applying the formula from this section gives,

\[{x^2} + 4 = - 1\hspace{0.25in}{\mbox{or}}\hspace{0.25in}{x^2} + 4 = 1\] Show Step 2

To finish this problem all we need to do is solve each of the quadratic equations we got in the previous step. Here is the solution to each of them.

\[\begin{align*}{x^2} + 4 & = - 1\hspace{0.25in} \to \hspace{0.25in}{x^2} = - 5\hspace{0.25in} \to \hspace{0.25in}x = \pm \sqrt 5 \,i\\ {x^2} + 4 & = 1\hspace{0.25in} \to \hspace{0.25in}{x^2} = - 3\hspace{0.25in} \to \hspace{0.25in}x = \pm \sqrt 3 \,i\end{align*}\]

Note that the instructions asked for “real valued solutions”. This basically means that we don’t want complex solutions and the solutions to both of the quadratic equations from the first step are complex and so, for this equation, there are no solutions.