Factorisation and Completing the Square

**Nick87** · February 25th 2008, 04:02 PM

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.

1. Solve quadratic by factorising:

$y^2+8y=0$

2. Solve by completing the square:

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

Thanks in advance!

**topsquark** · February 25th 2008, 04:48 PM

Originally Posted by Nick87

1. Solve quadratic by factorising:

$y^2+8y=0$

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

**Krizalid** · February 25th 2008, 04:59 PM

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.

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