# Math Help - Quadratics: Completing the square

1. ## Quadratics: Completing the square

Hi, I've been doing some GCSE maths revision - working on completing the square. I was going through questions like x²+4x-3=0 using (x+a)+b and then solving them. However, I reached questions with numbers in front of the x² and couldn't understand how to solve it:

2x²-6x+1=0

2. Hello,

Factorise by 2 and then do as usual

3. I don't understand - how can I factorise it by 2 when there are no numbers that multiply to give one and add to give -6? Sorry!

4. Originally Posted by Turple
Hi, I've been doing some GCSE maths revision - working on completing the square. I was going through questions like x²+4x-3=0 using (x+a)+b and then solving them. However, I reached questions with numbers in front of the x² and couldn't understand how to solve it:

2x²-6x+1=0

$2x^2-6x+1=0$

Let's take that two out.

$2(x^2 - 3x) + 1 = 0$

Complete the square.

In this bracket, $(x^2 - 3x)$ , we want to add $+2,25$ inside there.
Why? Because the co-efficient of the $-3x$ could only have been possible when we had $2 \times -1,5$.
And: $(-1,5)^2 = +2,25$

But we cant forget about the 2 outside the bracket. $2(x^2 - 3x)$

So now we can say this:

$2(x^2 - 3x) + 2(2,25) - 2(2,25) + 1 = 0$

$2(x^2 - 3x + 2,25) - 4,5 + 1 = 0$

$2(x - 1,5)^2 - 3,5 = 0$

5. Oh i see, sorry ^^

$2x^2-6x+1=2(x^2-3x+\frac{1}{2})=2(x^2-2*\frac{3}{2} x+\frac{1}{2})$

$(\frac{3}{2})^2=\frac{9}{4}$

$2(\underbrace{x^2-2*\frac{3}{2} x+\frac{9}{4}}_{(x-\frac{3}{2})^2}-\frac{9}{4}+\frac{1}{2})=2(x-\frac{3}{2})^2+2(\underbrace{-\frac{9}{4}+\frac{1}{2}}_{-\frac{7}{4}})$

So $2x^2-6x+1=2(x-\frac{3}{2})^2-\frac{7}{2}$

I haven't checked my calculus yet, but did you understand how to do ?

6. Originally Posted by Moo
Oh i see, sorry ^^

$2x^2-6x+1=2(x^2-3x+\frac{1}{2})=2(x^2-2*\frac{3}{2} x+\frac{1}{2})$

$(\frac{3}{2})^2=\frac{9}{4}$

$2(\underbrace{x^2-2*\frac{3}{2} x+\frac{9}{4}}_{(x-\frac{3}{2})^2}-\frac{9}{4}+\frac{1}{2})=2(x-\frac{3}{2})^2+2(\frac{-\frac{9}{4}+\frac{1}{2}}_{-\frac{7}{4}})$

So $2x^2-6x+1=2(x-\frac{3}{2})^2-\frac{7}{2}$

I haven't checked my calculus yet, but did you understand how to do ?
Beat you to it, Moo

7. I hate writing in latex

I'll get my revenge jan

8. Woah. Sorry but I am so confused XD I can't understand where the 2.25 comes from?

9. Originally Posted by Turple
Woah. Sorry but I am so confused XD I can't understand where the 2.25 comes from?
$2.25 = 1.5^2$

10. So where does that 1.5 come from?

Sorry :/

11. Originally Posted by Turple
So where does that 1.5 come from?

Sorry :/
$(x - 1.5)^2 = x^2 - 3x + 2.25$ (Eugh, I hate typing a dot for the decimal seperator)

Do you see now?

12. See my demo, there are fractions and no questioning about "where does it come"

Ok ok i'm leaving ^^

13. So there is an x² and a -3x? Has the orginial quadratic been halved? And if it has I'm lost with that 2.25 again...

14. Originally Posted by Moo
Ok ok i'm leaving ^^
I'm thinking you should rather take over...

15. Ok Moo I will try to understand yours XD

Sorry to both fo you if I'm being frustratingly thick :/

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