# Thread: Factorisation and Completing the Square

1. ## Factorisation and Completing the Square

Maths is certainly not me, why on earth I decided to pick Further is beyond comprehension. A couple of questions have got me, eternal thanks for any response.

$y^2+8y=0$

2. Solve by completing the square:

$3x^2-5x-1=0$

2. Originally Posted by Nick87

$y^2+8y=0$
What factor is common to both terms?

Originally Posted by Nick87
2. Solve by completing the square:

$3x^2-5x-1=0$
$3x^2 - 5x - 1 = 0$

$x^2 - \frac{5}{3}x - \frac{1}{3} = 0$

$x^2 - \frac{5}{3}x = \frac{1}{3}$

Now, you have this equation and the left hand side is in the form of
$x^2 + 2bx$

By comparison we get that $2b = -\frac{5}{3} \implies b = -\frac{5}{6}$.

We want to add something to $x^2 + 2bx$ that will make it a perfect square. If you look at the formula $(a + b)^2 = a^2 + 2ab + b^2$ you will see that if we add a $b^2$ to the two terms we get
$x^2 + 2bx + b^2 = (x + b)^2$

So we want to add a $b^2 = \frac{25}{36}$ to both sides of the equation:
$x^2 - \frac{5}{3}x + \frac{25}{36} = \frac{1}{3} + \frac{25}{36}$

So...

$\left ( x - \frac{5}{6} \right ) ^2 = \frac{37}{36}$

Can you finish it from here?

-Dan

3. Or for the second one:

\begin{aligned}3x^2 - 5x - 1 &=0\\ 36x^2 - 60x - 12 &=0\\ (6x - 5)^2 &= 37.\end{aligned}

Dan's answer is in agreement with mine.