Chapter 12.5, Problem 14WE

To determine

Solve the quadratic equation.

Answer to Problem 14WE

  $x=4\pm 3\sqrt{3}$

Explanation of Solution

Given:

The equation is :

  $x^2-8x=11$

Calculation:

The given equation $x^2-8x=11\Rightarrow x^2-8x-11=0$ is of the type $x^2+bx+c=0$ , where b = -8 is even .

So, it can be solved by Completing the square.

So,

  $\begin{array}{l}{x}^2-8x=11\\ \Rightarrow x^2-8x+16=11+\mathrm{16.........}[add{\left(\frac{-8}{2}\right)}^2=16\text{ each side]}\\ \Rightarrow {\left(x-4\right)}^2=27\\ \Rightarrow x-4=\pm \sqrt{27}\mathrm{...................}[\text{take square root each side]}\\ \Rightarrow x=4\pm 3\sqrt{3}\mathrm{.....................}[\text{add 4 each side]}\end{array}$