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Is it possible to form a quadratic simultaneous equation/linear with both solutions?

Is there a method that finds the equations of a quadratic simultaneous equation and a linear equation when you know that the solutions are (16/13,1/3) and (-4,5)

?
Thinking more generally would be beneficial.

You are asking;

Given points (s,t) and (u,v) is there a

quadratic

y = ax^2 + bx + c

and a line

y = dx + e

such that these points are on the line and quadratic?
Reply 2
Original post by DeanK22
Thinking more generally would be beneficial.

You are asking;

Given points (s,t) and (u,v) is there a

quadratic

y = ax^2 + bx + c

and a line

y = dx + e

such that these points are on the line and quadratic?


Yes this.

How would you go about finding the exact equations that satisfy both sets of solutions?
Reply 3
So I've just uncovered a part of maths no one knows anything about? Hmm...

Surely if you can get the solutions, you get get the equations.... Just like find the equation of a circle
Original post by Student#254
So I've just uncovered a part of maths no one knows anything about?


What are you talking about ...

I was giving a hint.

Do you think that given two points there is a line through those points?

Given two points do you think there is a quadratic through those points?
Reply 5
Original post by DeanK22
What are you talking about ...

I was giving a hint.

Do you think that given two points there is a line through those points?

Given two points do you think there is a quadratic through those points?


Oh ffs... Please stop with the 'do you think'...

I want answers now please. I don't do these ridiculous mind games...
Reply 6
Original post by Student#254
Oh ffs... Please stop with the 'do you think'...

I want answers now please. I don't do these ridiculous mind games...


The answers are pretty obvious to the posed questions. . .
Reply 7
Original post by Student#254
Oh ffs... Please stop with the 'do you think'...

I want answers now please. I don't do these ridiculous mind games...


Maybe the benefit is that you have to think these things out for yourself to truly gain from the question? Do you think that's right?
Reply 8
Original post by jaheen22
The answers are pretty obvious to the posed questions. . .


Original post by davidmarsh01
Maybe the benefit is that you have to think these things out for yourself to truly gain from the question? Do you think that's right?


No they are not. Well not to me anyway...

I see an almost infinite set of equations that could satisfy those solutions...

x^2+bx+c
y=dx+e

I see it like this...

If b increases/decreases then d would change. They're all variables so how do you calculate the equations?

Are there more than two outcomes? Basic example...

x^2+2x+1 and y=2x+1 does satisfy both solutions (not accurate)

while...

x^2+4x+1 and y=x+1 also satisfies both equations...

Do you see what I mean here?
This problem shouldn't be too difficult if you ignore this business with variables - it can lead to a correct answer though with some algebraic (tedious algebraic) effort.

Taking any two points there is certainly a (unique) line that passes through them.

So we need not concern ourselves with worrying about this line. So all that matters in reality - given two points is there a quadratic containing the two points.

I would recommend some trial cases with a couple of pairs of points.

Spoiler

Reply 11
Original post by DeanK22

Spoiler



I don't take a maths degree so fyi I'm not that great at maths... All I wanted was help to find the equations...but no I got responses that really made no sense to me at all.

I need concise explanations for the not so mathematically inclined.

I appreciate your help but it's confusing me :confused:
For example could you find a quadratic through

(1,2) and (1,-2) ?

Or (1,1) , (2,2) and (2,3) ?

Spoiler



So your answer is in large - yes you can - but in general no you can't.

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