Chapter 7.1, Problem 13WE

To determine

To Solve:

  $x^2-2x-5=0$ by completing the square.

Answer to Problem 13WE

  $x=1\pm \sqrt{6}$

Explanation of Solution

Given:

The equation is :

  $x^2-2x-5=0$

Calculation:

Given:

  $\begin{array}{l}{x}^2-2x-5=0\\ \Rightarrow x^2-2x=\mathrm{5................}[\text{add 5 each side to isolate the constant]}\\ \Rightarrow x^2-2x+{\left(\frac{-2}{2}\right)}^2=5+{\left(\frac{-2}{2}\right)}^2\mathrm{....}\left[\begin{array}{l}\text{add the square of the half of the coefficient}\\ \text{of x each side}\end{array}\right]\\ \Rightarrow x^2-2x+1=5+1\\ \Rightarrow {\left(x-1\right)}^2=\mathrm{6.................................}[{\left(a-b\right)}^2=a^2-2ab+b^2]\\ \Rightarrow x-1=\pm \sqrt{6}\mathrm{.................................}[\text{take square root each side]}\\ \Rightarrow x=1\pm \sqrt{6}\mathrm{...................................}[\text{add 1 each side]}\end{array}$

Since 6 does not have any perfect square root, it is in the simplest form.

So,

  $x=1\pm \sqrt{6}$