Chapter 12.2, Problem 1WE

To determine

To Solve: The equation $x^2-4x=17$ by completing the square.

Answer to Problem 1WE

The required solution set is $\left\{2+\sqrt{21},2-\sqrt{21}\right\}$ .

Explanation of Solution

Given information:

  $x^2-4x=17$

Calculation:

  $\begin{array}{l}{x}^2-4x=17\\ x^2-4x+{\left(-\frac{4}{2}\right)}^2=17+{\left(-\frac{4}{2}\right)}^2\end{array}$

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

  $\begin{array}{l}{x}^2-4x+{\left(-2\right)}^2=17+{\left(-2\right)}^2\\ x^2-4x+4=17+4\\ {\left(x-2\right)}^2=21\end{array}$

(Simplify)

  $\begin{array}{l}{\left(x-2\right)}^2=21\\ x-2=\pm \sqrt{21}\end{array}$

(Take square root on both sides)

  $x=2\pm \sqrt{21}$

(Add 2 to each side)

  $\begin{array}{l}x=2+\sqrt{21}\\ x=6.6\end{array}$

Or

  $\begin{array}{l}x=2-\sqrt{21}\\ x=-2.6\end{array}$

(Round off to nearest tenth)

Therefore, the solution set is $\left\{2+\sqrt{21},2-\sqrt{21}\right\}$ .