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 $$ (x+1)(x-6). $$