this quadratic equation doesn't seem to factorise

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October 12th 2009, 01:10 PM
portia

this quadratic equation doesn't seem to factorise

Hi guys,

I've been trying to factorise the following equation:

4p^2 + 12p + 9 = 0

I can't seem to find any values for a and b so that 4a+b=12 and a*b=9

Any help on this?

thanks
October 12th 2009, 01:21 PM
artvandalay11

Quote:

Originally Posted by portia

Hi guys,

I've been trying to factorise the following equation:

4p^2 + 12p + 9 = 0

I can't seem to find any values for a and b so that 4a+b=12 and a*b=9

Any help on this?

thanks

(2p+3)(2p+3)
October 12th 2009, 01:25 PM
portia

Thanks, I do appreciate your answer, however, I'd like to know how you did it:)
October 12th 2009, 01:32 PM
artvandalay11

Quote:

Originally Posted by portia

Thanks, I do appreciate your answer, however, I'd like to know how you did it:)

I find that explaining methods of factoring is very tricky... basically I take advantage of the fact that I had it drilled into me and I am quite quick with multiplying in my head

But here's my shot at it

$4x^2+12x+9$

4 has the following factors 1,2,2,4

So if this thing is factorable it has one of the two forms below

1) (x )(4x ) or
2) (2x )(2x )

So I just pick one and run with it, let's say we guess wrong

Try (x )(4x )

Now 9 has the factors 1,3,3,9 so we can put 1,9 together in any order or 3,3 together

Since 9 is positive, either both numbers are negative or both numbers are positive

So our possible combinations are: (x+3)(4x+3), (x-3)(4x-3), (x+1)(4x+9), (x+9)(4x+1), (x-1)(4x-9), (x-9)(4x-1)

I run through all these in my head and realize that I was not getting 12x out of the deal, so I move to the next one

Try (2x )(2x )

Now for blank space we have exactly the same combinations as above, I run through them in my head (and by this I mean I say to myself (2x+1)(2x+9) will give me 20x, nope no good try again)

and I get that (2x+3)(2x+3) is my answer

hope this helps

and by the way, this method is significantly quicker than setting up those equations and looking at it once you get good with this
October 12th 2009, 01:49 PM
portia

Ok, I kind of understand it. I also used the quadratic formula and got -3/2 which makes sense:

(x+3/2)(x+3/2) /// multiplied by 2

(2x+3)(2x+3)

So in this case there's only one solution: x=-3/2

I'll need to practise it a lot more. I am ok when the coefficient 'a' is 1 as in x^2+4x........ but when 'a' is different to 1 then I sometimes struggle.

thanks
October 12th 2009, 01:51 PM
artvandalay11

you can of course always apply the quadratic formula

When you apply it, you get 2 answers in general, call them $r_1$ and $r_2$

Then the quadratic can be written as $(x-r_1)(x-r_2)$

In your case $r_1=r_2=-\frac{3}{2}$

So if you were stuck on a test and had time this is a fail safe, but obviously factoring is preferred
October 12th 2009, 01:56 PM
e^(i*pi)

The quick way to see whether a quadratic will factorise (at least until you're good doing in mentally) is to find the discriminant, $\Delta$

$\Delta = b^2-4ac$

If $\Delta$ is a perfect square it will factorise
October 12th 2009, 01:58 PM
artvandalay11

yeah but you can end up with fractions as the factors in which case I would just use the quadratic formula anyway
October 12th 2009, 02:29 PM
Barthayn

Quote:

Originally Posted by portia

Hi guys,

I've been trying to factorise the following equation:

4p^2 + 12p + 9 = 0

I can't seem to find any values for a and b so that 4a+b=12 and a*b=9

Any help on this?

thanks

this is a complex trinomal, this means you have to multiply the A value with the C value and find two numbers that add up to 12 and multiply to 36. In this case it will be 6p and 6p. From there to find the great common factor between the first two values and the next two values. If you did it right you will get one value twice in your work.

This is how your work should work:
$0 = 4p^2+12p+9$
$0 = 4p^2+6p+6p+9$
$0 = 2p(2p+3)+3(2p+3)$
$0 = (2p+3)(2p+3)$
$0 = (2p+3)^2$

Therefore $(2p+3)^2$ is $4p^2+12p+9=0$ .

P cannot equal -3 over 2. The root is -3 over 2.