Chapter 12.5, Problem 15WE

To determine

Solve the quadratic equation.

Answer to Problem 15WE

  $x=\frac{1\pm \sqrt{22}}{3}$

Explanation of Solution

Given:

The equation is :

  $3x^2=2x+7$

Calculation:

The given equation $3x^2=2x+7\Rightarrow 3x^2-2x-7=0$ is of the type $a{x}^2+bx+c=0$ , where a = 3 and b = -2, c = -7 .

  $b^2-4ac={(-2)}^2-4(3)(-7)=88$ , which is not a perfect square .

So, it can be solved by using Quadratic Formula:

So,

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-(-2)\pm \sqrt{88}}{2(3)}\\ \Rightarrow x=\frac{2\pm 2\sqrt{22}}{2(3)}\\ \Rightarrow x=\frac{1\pm \sqrt{22}}{3}\end{array}$