Chapter 12.3, Problem 13WE

To determine

To solve: the equation $-4x^2+2x+3=0$ using quadratic formula. Give irrational roots in simplest radical form and then approximate them to the nearest tenth.

Answer to Problem 13WE

The required answer is: solution set is - 0.65, 1.15

Explanation of Solution

Given information:

The given equation: $-4x^2+2x+3=0$

Calculation:

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

(Quadratic formula)

Where $a=-4,b=2andc=3$

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

(Remove the brackets)

  $\begin{array}{l}=\frac{-2\pm \sqrt{4+48}}{-8}\\ =\frac{-2\pm \sqrt{52}}{-8}\\ =\frac{-2\pm \sqrt{4\cdot 13}}{-8}\end{array}$

(Product property of square roots)

  $\begin{array}{l}=\frac{-2\pm 2\sqrt{13}}{-8}\\ =\frac{2(-1\pm \sqrt{13})}{-8}\end{array}$

(Distributive property)

  $=\frac{-1\pm \sqrt{13}}{-4}$

(Simplify)

Since $\sqrt{13}\approx 3.61$ , $x\approx \frac{-1+3.61}{-4}=\frac{2.61}{-4}\approx -0.65$

Or     $x\approx \frac{-1-3.61}{-4}=\frac{-4.61}{-4}\approx 1.15$

Therefore the solution set is $\frac{-1+\sqrt{13}}{-4},\frac{-1-\sqrt{13}}{-4}$ or - 0.65, 1.15