To complete the square, remember the formula for a square binomial:
[tex](x+a)^2=x^2+2ax+a^2[/tex]
The coefficient of the linear term is 2a. In this case:
[tex]x^2-x=4[/tex]
We can see that the coefficient of the linear term on the left member of the equation is -1. Then:
[tex]\begin{gathered} 2a=-1 \\ \Rightarrow a=-\frac{1}{2} \\ \Rightarrow a^2=\frac{1}{4} \end{gathered}[/tex]
Add and substract 1/4 to the left member of the equation:
[tex]x^2-x+\frac{1}{4}-\frac{1}{4}=4[/tex]
Since the first three terms of the left member correspond to a perfect square trinomial, then we can rewrite it as a square binomial:
[tex]\begin{gathered} x^2-x+\frac{1}{4}=(x-\frac{1}{2})^2 \\ \Rightarrow(x-\frac{1}{2})^2-\frac{1}{4}=4 \end{gathered}[/tex]
Add 1/4 to both sides of the equation:
[tex]\begin{gathered} \Rightarrow(x-\frac{1}{2})^2=4+\frac{1}{4} \\ \Rightarrow(x-\frac{1}{2})^2=\frac{17}{4} \end{gathered}[/tex]
Take the square root to both sides of the equation:
[tex]\begin{gathered} \Rightarrow\sqrt[]{(x-\frac{1}{2})^2}=\sqrt[]{\frac{17}{4}} \\ \Rightarrow x-\frac{1}{2}=\pm\frac{\sqrt[]{17}}{2} \\ \Rightarrow x=\pm\frac{\sqrt[]{17}}{2}+\frac{1}{2} \end{gathered}[/tex]
Therefore, the answer is:
[tex]x=\pm\frac{\sqrt[]{17}}{2}+\frac{1}{2}[/tex]
