Chapter 8.1, Problem 9MRE

To determine

To solve the equation over real numbers.

Answer to Problem 9MRE

The solution of the equation is $\frac{1\pm 2\sqrt{3}}{3}$ .

Explanation of Solution

Given information:

The given equation is:

  ${\left(3k-1\right)}^2=12$

Formula used:

The following formula is used:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation :

The solution can be obtained by:

  $\begin{array}{l}{\left(3k-1\right)}^2=12\\ 9k^2-6k+1=12\left[\text{Evaluate the square of the expression}\right]\\ 9k^2-6k+1-12=0\\ 9k^2-6k-11=0\left[\text{Write the equation in the form }a{x}^2+bx+c=0\right]\\ k=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ k=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(9\right)\left(-11\right)}}{2\left(9\right)}\\ k=\frac{6\pm \sqrt{36+396}}{18}\\ k=\frac{6\pm \sqrt{432}}{18}\\ k=\frac{6\pm 12\sqrt{3}}{18}\\ k=\frac{6\left(1\pm 2\sqrt{3}\right)}{18}\\ k=\frac{1\pm 2\sqrt{3}}{3}\end{array}$

Hence, the solution of the equation is $\frac{1\pm 2\sqrt{3}}{3}$ .