Chapter 12.3, Problem 8WE

To determine

To solve: the equation $k^2-3k-1=0$ using quadratic formula. Give irrational roots in simplest radical form and then approximate them to the nearest tenth.

Answer to Problem 8WE

The required answer is: solution set is 3.31, - 0.31

Explanation of Solution

Given information:

The given equation: $k^2-3k-1=0$

Calculation:

  $\begin{array}{l}{k}^2-3k-1=0\\ k=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$

(Quadratic formula)

Where $a=1,b=-3andc=-1$

  $k=\frac{-(-3)\pm \sqrt{{(-3)}^2-4(1)(-1)}}{2(1)}$

(Remove the brackets)

  $\begin{array}{l}=\frac{3\pm \sqrt{9+4}}{2}\\ =\frac{3\pm \sqrt{13}}{2}\end{array}$

(Simplify)

Since $\sqrt{13}\approx 3.61$ , $k\approx \frac{3+3.61}{2}=\frac{6.61}{2}\approx 3.31$

Or

  $k\approx \frac{3-3.61}{2}=\frac{-0.61}{2}\approx -0.31$

Therefore the solution set is $\frac{3+\sqrt{13}}{2},\frac{3-\sqrt{13}}{2}$ or 3.31, - 0.31