Chapter 12.5, Problem 2WE

To determine

Solve the quadratic equation.

Answer to Problem 2WE

  $x\in \left\{1+\sqrt{2},1-\sqrt{2}\right\}$

Explanation of Solution

Given:

The equation is :

  $x^2-2x=1$

Calculation:

The given equation $x^2-2x=1\Rightarrow x^2-2x-1=0$ is of the type $x^2+bx+c=0$ , where b = -2 is even .

So, it can be solved by Completing the square.

  $\begin{array}{l}{x}^2-2x=1\\ \Rightarrow x^2-2x+1=1+\mathrm{1.............}[add{\left(\frac{-2}{2}\right)}^2=1\text{ each side]}\\ \Rightarrow {\left(x-1\right)}^2=2\\ \Rightarrow x-1=\pm \sqrt{2}\mathrm{.....................}[\text{take square root each side]}\\ \Rightarrow x=1\pm \sqrt{2}\mathrm{.......................}[add\text{ 1 each side]}\end{array}$

  $x\in \left\{1+\sqrt{2},1-\sqrt{2}\right\}$