Chapter 12.2, Problem 19WE

To determine

To Solve: The equation $\frac{x^2}{4}-x=3$ and write irrational roots in simplest radical form.

Answer to Problem 19WE

The required solution set is $\left\{6,-2\right\}$ .

Explanation of Solution

Given information:

  $\frac{x^2}{4}-x=3$

Calculation:

  $\begin{array}{l}\frac{x^2}{4}-x=3\\ x^2-4x=12\end{array}$

(Multiply both sides by 4 so that the coefficient of $x^2$ will be 1)

  $x^2-4x+{\left(-\frac{4}{2}\right)}^2=12+{\left(-\frac{4}{2}\right)}^2$

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

  $\begin{array}{l}{x}^2-4x+{\left(2\right)}^2=12+{\left(2\right)}^2\\ x^2-4x+4=12+4\\ {\left(x-2\right)}^2=16\end{array}$

(Simplify)

  $\begin{array}{l}x-2=\sqrt{16}\\ x-2=\pm 4\end{array}$

(Perform square root on both sides)

  $x=2\pm 4$

(Add 2 to each side)

  $\begin{array}{l}x=2+4\\ x=6\end{array}$

Or

  $\begin{array}{l}x=2-4\\ x=-2\end{array}$

Therefore, the solution set is $\left\{6,-2\right\}$