Chapter 12.2, Problem 8WE

To determine

To Solve: The equation

  $2b^2+16b=4$ by completing the square.

Answer to Problem 8WE

The required solution set is $\left\{-4+3\sqrt{2},-4-3\sqrt{2}\right\}$ .

Explanation of Solution

Given information:

  $2b^2+16b=4$

Calculation:

  $\begin{array}{l}2b^2+16b=4\\ b^2+8b=2\end{array}$

(Divide both sides by 2 so that the coefficient of $x^2$ will be 1)

  $b^2+8b+{\left(\frac{8}{2}\right)}^2=2+{\left(\frac{8}{2}\right)}^2$

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

  $\begin{array}{l}{b}^2+8b+{\left(4\right)}^2=2+{\left(4\right)}^2\\ b^2+8b+16=2+16\\ {\left(b+4\right)}^2=18\end{array}$

(Simplify)

  $b+4=\pm \sqrt{18}$

(Take square root on each side)

  $b+4=\pm 3\sqrt{2}$

(Simplify)

  $b=-4+3\sqrt{2}$

Or

  $b=-4-3\sqrt{2}$

Therefore, the solution set is $\left\{-4+3\sqrt{2},-4-3\sqrt{2}\right\}$

Now round off the solution to its nearest tenth

  $\begin{array}{l}b=-4+3\sqrt{2}\\ b=0.2\end{array}$

Or

  $\begin{array}{l}b=-4-3\sqrt{2}\\ b=-8.2\end{array}$