# Looking for a quadratic that uses the formula to solve.Watch

Thread starter 3 years ago
#1
Howdy ,

Looking for a few examples of quadratic equations that need .

I have been able to solve most by factorising.

Just wanted to see when it would be more beneficial to use the attached formula to solve.

Thanks
0
3 years ago
#2
The formula is useful where the expression will not factorise which will be the case when b^2-4ac is not a perfect square. If you randomly write a quadratic, making sure c is negative and a and b are positive then most of them will not factorise and you will have to use the formula.

If you are not careful with signs then b^2-4ac <0 and there is no real solution
0
3 years ago
#3
Easy to make up some of your own as above, as long as b^2 > 4ac (otherwise you'd be taking the square root of a negative number, meaning no real rools).

Eg b=3, a = c = 1.

x^2 + 3x + 1 (which can't be factorised)

b =6, a=2, c=3.

2x^2 + 6x + 3 (which can't be factorised) and so on.
1
3 years ago
#4
If you don't see how to factorise it, you should either complete the square or use the formula. If the roots aren't rational, then you usually need to.

(Original post by nerak99)
The formula is useful where the expression will not factorise which will be the case when b^2-4ac is not a perfect square. If you randomly write a quadratic, making sure c is negative and a and b are positive then most of them will not factorise and you will have to use the formula.

If you are not careful with signs then b^2-4ac <0 and there is no real solution
All quadratics factorise. You just have to work in a big enough number system.
1
3 years ago
#5
(Original post by morgan8002)
All quadratics factorise. You just have to work in a big enough number system.
True and also massively helpful in the context of the OP

Edit: "All" really?
0
3 years ago
#6
(Original post by nerak99)
"All" really?
Yes.
0
3 years ago
#7
(Original post by nerak99)
Edit: "All" really?
D'you have a counter example?
0
3 years ago
#8
Well, starting from the fact that this did not start with me making any claim.

To find the appropriate system you would need to solve the quadratic in order to find it. Maybe I am wrong there but it has to work like this, I find a non factorisable quadratic which you then have to find the appropriate system without, effectively, solving the equation.

Otherwise you can always just write (x- one root from formula)(x- other root)

I guess I didn't realise just how trivial the claim was as well as utterly irrelevant to the OP.
0
Thread starter 3 years ago
#9
So can all quadratics be factorised?
0
3 years ago
#10
(Original post by makin)
So can all quadratics be factorised?
no, that's why we have the quadratic formula
0
3 years ago
#11
(Original post by thefatone)
no, that's why we have the quadratic formula
...uhm, that's exactly what makes them factorisable?
0
3 years ago
#12
(Original post by Zacken)
...uhm, that's exactly what makes them factorisable?
but what if the curve doesn't touch the x-axis?
0
3 years ago
#13
(Original post by thefatone)
no, that's why we have the quadratic formula
The quadratic formula is technically a factorisation you obtain from completing the square. 0
3 years ago
#14
(Original post by thefatone)
but what if the curve doesn't touch the x-axis?
Complex.
0
3 years ago
#15
(Original post by Zacken)
Complex.
complex? things happen xD
0
3 years ago
#16
(Original post by thefatone)
complex? things happen xD
I'm guessing you haven't heard of the complex field which is the algebraic extension of ? 0
3 years ago
#17
(Original post by Zacken)
I'm guessing you haven't heard of the complex field which is the algebraic extension of ? nope but i have seen that r before.

it was in a question which said x can be any real number and it had that r in the notation that's about as much as i know xD
0
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