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Solve (u + 2) ^ 2 - 8= 0 , where u is a real number. Simplify your answer as much as possible . If there is more than one solution, separate them with commas. If there is no solution, click "No solution."

1 Answer

5 votes

The equation is given as


(u+2)^2-8=0

Before we begin solving, we need to expand the equation:


\begin{gathered} (u+2)(u+2)-8=0 \\ u^2+2x+2x+4-8=0 \\ u^2+4x-4=0 \end{gathered}

We can solve the equation using the quadratic formula:


x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

where


\begin{gathered} a=1 \\ b=4 \\ c=-4 \end{gathered}

Substituting, we have


\begin{gathered} u=\frac{-4\pm\sqrt[]{4^2-(4*1*-4)}}{2*1} \\ u=\frac{-4\pm\sqrt[]{32}}{2} \end{gathered}

Therefore, we can calculate the values of u to be


\begin{gathered} u=\frac{-4+\sqrt[]{32}}{2} \\ u=-2+2\sqrt[]{2} \end{gathered}

or


\begin{gathered} u=\frac{-4-\sqrt[]{32}}{2} \\ u=-2-2\sqrt[]{2} \end{gathered}

Therefore, the roots are given as


u=\mleft\lbrace-2+2\sqrt[]{2},-2-2\sqrt[]{2}\mright\rbrace

User CyclingDave
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