0
$\begingroup$

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}

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thanks for the reply. It was actually an interactive question from Khan Academy that was coming back with incorrect answer when I filled out the boxes but I've just realised one of the boxes was missing a negative sign and thats why it wasn't taking it! I thought I had messed up along the fractions somewhere. Your answer has helped cement the concept further though. Thanks! $\endgroup$ – Anonan Oct 19 '15 at 21:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.