Chapter 12.3, Problem 18WE

To determine

To solve: the equation $2r=5-4r^2$ using quadratic formula. Give irrational roots in simplest radical form and then approximate them to the nearest tenth.

Answer to Problem 18WE

The required answer is: solution set is 0.9, - 1.4

Explanation of Solution

Given information:

The given equation: $2r=5-4r^2$

Calculation:

  $\begin{array}{l}2r=5-4r^2\\ 2r+4r^2=5-4r^2+4r^2\end{array}$

  $(Add4r^{2}tobothsides)$

  $\begin{array}{l}2r+4r^2=5\\ 2r+4r^2-5=5-5\end{array}$

(Subtract 5 from both sides)

  $\begin{array}{l}2r+4r^2-5=0\\ 4r^2+2r-5=0\end{array}$

(Simplify)

  $r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

(Quadratic formula)

Where $a=4,b=2andc=-5$

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

(Remove the brackets)

  $\begin{array}{l}=\frac{-2\pm \sqrt{4+80}}{8}\\ =\frac{-2\pm \sqrt{84}}{8}\\ =\frac{-2\pm \sqrt{4\cdot 21}}{8}\end{array}$

(Product property of square roots)

  $\begin{array}{l}=\frac{-2\pm 2\sqrt{21}}{8}\\ =\frac{2(-1\pm \sqrt{21})}{8}\end{array}$

(Distributive property)

  $=\frac{-1\pm \sqrt{21}}{4}$

(Simplify)

Since $\sqrt{21}\approx 4.58$ , $r\approx \frac{-1+4.58}{4}=\frac{3.58}{4}\approx 0.9$

Or   $r\approx \frac{-1-4.58}{4}=\frac{-5.58}{4}\approx -1.4$

Therefore the solution set is $\frac{-1+\sqrt{21}}{4},\frac{-1-\sqrt{21}}{4}$ or 0.9, - 1.4