Chapter 12.3, Problem 10WE

To determine

To solve: the equation $n^2-6n-1=0$ using quadratic formula. Give irrational roots in simplest radical form and then approximate them to the nearest tenth.

Answer to Problem 10WE

The required answer is: solution set is 6.16, - 0.16

Explanation of Solution

Given information:

The given equation: $n^2-6n-1=0$

Calculation:

  $\begin{array}{l}{n}^2-6n-1=0\\ n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$

(Quadratic formula)

Where $a=1,b=-6andc=-1$

  $n=\frac{-(-6)\pm \sqrt{{(-6)}^2-4(1)(-1)}}{2(1)}$

(Remove the brackets)

  $\begin{array}{l}=\frac{6\pm \sqrt{36+4}}{2}\\ =\frac{6\pm \sqrt{40}}{2}\\ =\frac{6\pm \sqrt{4\cdot 10}}{2}\end{array}$

(Product property of square roots)

  $\begin{array}{l}=\frac{6\pm 2\sqrt{10}}{2}\\ =\frac{2(3\pm \sqrt{10})}{2}\end{array}$

(Distributive property)

  $=3\pm \sqrt{10}$

(Simplify)

Since $\sqrt{10}\approx 3.16$ , $n\approx 3+3.16\approx 6.16$

Or

  $n\approx 3-3.16\approx -0.16$

Therefore the solution set is $3+\sqrt{10},3-\sqrt{10}$ or 6.16, - 0.16