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### Section 2.14 : Absolute Value Equations

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

$\left| {{x^2} + 2x} \right| = 15$

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Hint : Don’t let the fact that there is a quadratic term in the absolute value throw you off. This problem works exactly the same as the previous problems!
Start Solution

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,

${x^2} + 2x = - 15\hspace{0.25in}{\mbox{or}}\hspace{0.25in}{x^2} + 2x = 15$ 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 the first one given above.

\begin{align*}{x^2} + 2x & = - 15\\ {x^2} + 2x + 15 & = 0\hspace{0.25in} \to \hspace{0.25in}x = \frac{{ - 2 \pm \sqrt {4 - 4\left( 1 \right)\left( {15} \right)} }}{{2\left( 1 \right)}} = \frac{{ - 2 \pm \sqrt {56} \,i}}{2}\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 the first quadratic are clearly complex and so we won’t use them in our solution.

The solution to the second quadratic is,

\begin{align*}{x^2} + 2x - 15 & = 0\\ \left( {x + 5} \right)\left( {x - 3} \right) & = 0\hspace{0.25in} \to \hspace{0.25in}x = - 5,\,\,\,x = 3\end{align*}

Both of these are real solutions and so are acceptable solutions.

Therefore, the two solutions to the original equation are then : $$\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 5\,\,\,\,\,{\mbox{and}}\,\,\,\,\,x = 3}}$$ .