Chapter 4.2, Problem 16E

To determine

To solve: $3g^2-12g=-4$ by completing the square.

Answer to Problem 16E

  $g_{1,2}=2\pm \left(2\sqrt{2}\right)/\sqrt{3}$

Explanation of Solution

Given information: $3g^2-12g=-4$

Calculation:

We’ll have to complete the expression from the left side, to create a perfect square.

  $3g^2-12g=-4$

We’ll be guided by the formula that gives the perfect square.

  ${\left(a+b\right)}^2=a^2+2ab+b^2$

We’ve divided both sides the equation. Now it’s much more easier to complete the square from the left.

  $g^2-4g=-4/3$

We’ll identify the missing term that completes the square.

  $a^2+2ab+b^2=g^2-4g+b^2$

The term that completes the square is 4.

  $\begin{array}{l}2ab=-4g\\ \text{But }a=g\\ 2gb=-4g\Rightarrow b=-2\Rightarrow b^2=4\end{array}$

We’ll add both sides the missing term, to keep the equality.

  $g^2-4g+4=-4/3+4$

We’ll combine the terms from the right side.

  $\begin{array}{l}{\left(g-2\right)}^2=8/3\\ \left(g-2\right)=\pm \sqrt{8/3}\\ g_{1,2}=2\pm \left(2\sqrt{2}\right)/\sqrt{3}\end{array}$