Chapter 12.3, Problem 11WE

To determine

To solve: the equation $7x^2+2x-2=0$ using quadratic formula. Give irrational roots in simplest radical form and then approximate them to the nearest tenth.

Answer to Problem 11WE

The required answer is: solution set is 0.41, - 0.7

Explanation of Solution

Given information:

The given equation: $7x^2+2x-2=0$

Calculation:

  $\begin{array}{l}7x^2+2x-2=0\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$

(Quadratic formula)

Where $a=7,b=2andc=-2$

  $x=\frac{-(2)\pm \sqrt{{(2)}^2-4(7)(-2)}}{2(7)}$

(Remove the brackets)

  $\begin{array}{l}=\frac{-2\pm \sqrt{4+56}}{14}\\ =\frac{-2\pm \sqrt{60}}{14}\\ =\frac{-2\pm \sqrt{4\cdot 15}}{14}\end{array}$

(Product property of square roots)

  $\begin{array}{l}=\frac{-2\pm 2\sqrt{15}}{14}\\ =\frac{2(-1\pm \sqrt{15})}{14}\end{array}$

(Distributive property)

  $=\frac{-1\pm \sqrt{15}}{7}$

(Simplify)

Since $\sqrt{15}\approx 3.87$ , $x\approx \frac{-1+3.87}{7}=\frac{2.87}{7}\approx 0.41$

Or     $x\approx \frac{-1-3.87}{7}=\frac{-4.87}{7}\approx -0.7$

Therefore the solution set is $\frac{-1+\sqrt{15}}{7},\frac{-1-\sqrt{15}}{7}$ or 0.41, - 0.7