# Completing the Square (Solving for $x$ with fractions)

Im working through some completing the square questions but this question I'm currently working on has a few fractions which seem to be tripping me up. Some advice on where I have gone wrong would be greatly appreciated.

Q) Complete the Square to solve for $x$: $\qquad x^2 - 5x - 6 = 0$

So far I have:
\begin{align} x^2 - 5x &= 6 \\ x^2 - 5x + \frac{25}4 &= \frac{49}4 \\ \left(x - \frac52\right)\left(x - \frac52\right) &= \frac{49}4 \\ \left(x - \frac52\right)^2 = \frac{49}4 \end{align}

Take the square root:
\begin{align} x - \frac52 &= \pm \frac72 \\ x_1 &= \frac72 + \frac52 = \frac{12}2 = 6 \\ x_2 &= -\frac72 + \frac52 = -\frac22 = -1 \end{align}

What you have done is correct, $$0= x^2-5x-6 = \left( x-\frac{5}{2} \right)^2 - \frac{5^2}{2^2}-6 \\ = \left( x-\frac{5}{2} \right)^2 - \frac{49}{4},$$ so $$\left( x-\frac{5}{2} \right) = \pm \frac{7}{2},$$ so $x=-1$ or $x=6$. You can check these in the original equation: $$(-1)^2-5(-1)-6 = 0\\ 6^2-5 \times 6 -6 = 0.$$ Also, it is possible to spot the factorisation as $$(x+1)(x-6).$$