Chapter 12.2, Problem 10WE

To determine

To Solve: The equation $w^2+w=3$ by completing the square.

Answer to Problem 10WE

The required solution set is $\left\{\frac{\sqrt{13}-1}{2},\frac{-\sqrt{13}-1}{2}\right\}$ .

Explanation of Solution

Given information:

  $w^2+w=3$

Calculation:

  $\begin{array}{l}{w}^2+w=3\\ w^2+w+{\left(\frac{1}{2}\right)}^2=3+{\left(\frac{1}{2}\right)}^2\end{array}$

(Add square of half the coefficient of w to both sides)

  $\begin{array}{l}{w}^2+w+\left(\frac{1}{4}\right)=3+\left(\frac{1}{4}\right)\\ {\left(w+\frac{1}{2}\right)}^2=\frac{13}{4}\end{array}$

(Simplify)

  $\begin{array}{l}w+\frac{1}{2}=\sqrt{\frac{13}{4}}\\ w+\frac{1}{2}=\pm \frac{\sqrt{13}}{2}\end{array}$

(Perform square root on both sides)

  $w=\frac{\sqrt{13}-1}{2}$

Or

  $w=\frac{-\sqrt{13}-1}{2}$

Therefore, the solution set is $\left\{\frac{\sqrt{13}-1}{2},\frac{-\sqrt{13}-1}{2}\right\}$ .

Now round off the solution to its nearest tenth

  $\begin{array}{l}w=\frac{\sqrt{13}-1}{2}\\ w=1.3\end{array}$

Or

  $\begin{array}{l}w=\frac{-\sqrt{13}-1}{2}\\ w=-2.3\end{array}$