Chapter 12.5, Problem 12WE

To determine

Solve the quadratic equation.

Answer to Problem 12WE

  $x=\frac{1\pm \sqrt{13}}{4}$

Explanation of Solution

Given:

The equation is :

  $4x^2-2x-3=0$

Calculation:

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

  $b^2-4ac={(-2)}^2-4(4)(-3)=4+48=52$ , 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, $\begin{array}{l}x=\frac{-(-2)\pm \sqrt{52}}{2(4)}\\ \Rightarrow x=\frac{2\pm \sqrt{13\cdot 2^2}}{2\cdot 4}\\ \Rightarrow x=\frac{1\pm \sqrt{13}}{4}\end{array}$