Chapter 12.3, Problem 9WE

To determine

To solve: the equation $r^2+8r+5=0$ using quadratic formula. Give irrational roots in simplest radical form and then approximate them to the nearest tenth.

Answer to Problem 9WE

The required answer is: solution set is 0.68, - 7.32

Explanation of Solution

Given information:

The given equation: $r^2+8r+5=0$

Calculation:

  $\begin{array}{l}{r}^2+8r+5=0\\ r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$

(Quadratic formula)

Where $a=1,b=8andc=5$

  $r=\frac{-(8)\pm \sqrt{{(8)}^2-4(1)(5)}}{2(1)}$

(Remove the brackets)

  $\begin{array}{l}=\frac{-8\pm \sqrt{64-20}}{2}\\ =\frac{-8\pm \sqrt{44}}{2}\\ =\frac{-8\pm \sqrt{4\cdot 11}}{2}\end{array}$

(Product property of square roots)

  $\begin{array}{l}=\frac{-8\pm 2\sqrt{11}}{2}\\ =\frac{2(-4\pm \sqrt{11})}{2}\end{array}$

(Distributive property)

  $=-4\pm \sqrt{11}$

(Simplify)

Since $\sqrt{11}\approx 3.32$ , $r\approx -4+3.32\approx 0.68$

Or   $r\approx -4-3.32\approx -7.32$

Therefore the solution set is $-4+\sqrt{11},-4-\sqrt{11}$ or 0.68, - 7.32