Chapter 12.5, Problem 6WE

To determine

Solve the quadratic equation.

Answer to Problem 6WE

  $x=\frac{5\pm \sqrt{73}}{6}$

Explanation of Solution

Given:

The equation is :

  $3x^2-5x=4$

Calculation:

The given equation $3x^2-5x=4\Rightarrow 3x^2-5x-4=0$ is of the type $a{x}^2+bx+c=0$ , where a = 3 and b = -5, c = -4 .

  $b^2-4ac={(-5)}^2-4(3)(-4)=25+48=73$ , which is not perfect square .

So, it can be solved by using the Quadratic Formula : $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

So, $\begin{array}{l}x=\frac{-(-5)\pm \sqrt{73}}{2(3)}\\ \Rightarrow x=\frac{5\pm \sqrt{73}}{6}\end{array}$