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question

the root of ax^3+bx^2+cx+d is equal to the sum of its other two roots. Hence prove that b^3=4a(bc-2ad)

I realised the the roots must be equidistant ie the distance between them must be the same. I also though that a possibility of one root being 0 abd the other two equal in modulus. I was baffled after 15 minutes of trying. Can someone shed some light on this stuff.


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Avatar for DomStaff
DomStaff
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Original post by physicsmaths
the root of ax^3+bx^2+cx+d is equal to the sum of its other two roots. Hence prove that b^3=4a(bc-2ad)

I realised the the roots must be equidistant ie the distance between them must be the same. I also though that a possibility of one root being 0 abd the other two equal in modulus. I was baffled after 15 minutes of trying. Can someone shed some light on this stuff.


Posted from TSR Mobile


Not necessarily true.

Consider this:
(x−3)(x−5)(x−8)=x3−16x2+79x−120 (x-3)(x-5)(x-8)=x^3 -16x^2 +79x -120

Now these roots are not equidistant, are they? Their differences are 2 and 3. (From left to right). Yet the relationship of b3=4a(bc−2ad)b^3 = 4a(bc-2ad) still holds. Check it yourself on your calculator.

You can't consider that one root must be 0 in your answer (even though it may be in some cases), as I have just provided a counter-example of it having to be 0. Also, you are asked to prove it not to show it, so you can't just provide a special case and show it is true.
(edited 9 years ago)
Original post by DomStaff
Not necessarily true.

Consider this:
(x−3)(x−5)(x−8)=x3−16x2+79x−120 (x-3)(x-5)(x-8)=x^3 -16x^2 +79x -120

Now these roots are not equidistant, are they? Their differences are 2 and 3. (From left to right). Yet the relationship of b3=4a(bc−2ad)b^3 = 4a(bc-2ad) still holds. Check it yourself on your calculator.

You can't consider that one root must be 0 in your answer (even though it may be in some cases), as I have just provided a counter-example of it having to be 0. Also, you are asked to prove it not to show it, so you can't just provide a special case and show it is true.


so how would you prove it.


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Reply 4
Avatar for DomStaff
DomStaff
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Original post by physicsmaths
so how would you prove it.


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I'll have a go at it in a bit or tomorrow, but I would certainly use the 'roots of polynomials' aspect of A-Level maths to tackle it. If you haven't came across it, google it, if you have, then use it.
Original post by physicsmaths
so how would you prove it.


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Think about how you can obtain a relationship between the roots of your polynomial and it's coefficients
Reply 6
Avatar for DomStaff
DomStaff
2
I've not done it yet, but I've got the first few steps done in my head, ish. So when you get stuck I'll give you the first hint, which is how I will approach the problem when I come to do it.
lol everybody chill out i was making this horrible mistake time and time again. I shouldve got it right the first time. This is a question from 1968 btw. Suprised how i got deluded in some bogus method. Thanks for the help everyone. I cant believe i got this wrong first time.


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Original post by physicsmaths
the root of ax^3+bx^2+cx+d is equal to the sum of its other two roots. Hence prove that b^3=4a(bc-2ad)

I realised the the roots must be equidistant ie the distance between them must be the same. I also though that a possibility of one root being 0 abd the other two equal in modulus. I was baffled after 15 minutes of trying. Can someone shed some light on this stuff.


Posted from TSR Mobile


For a standard quadratic, you have α+β=−ba,αβ=ca\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a} where α,β\alpha, \beta are the roots.

There is a similar, but more involved, result that you can look up for the standard cubic. The result follows fairly easily once you know that result and you use the fact given in the question.
Reply 9
Avatar for DomStaff
DomStaff
2
Original post by physicsmaths
lol everybody chill out i was making this horrible mistake time and time again. I shouldve got it right the first time. This is a question from 1968 btw. Suprised how i got deluded in some bogus method. Thanks for the help everyone. I cant believe i got this wrong first time.


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So you've done it then, yes? I'll probably post my solution when I've done it anyways so that we can compare.
Original post by DomStaff
So you've done it then, yes? I'll probably post my solution when I've done it anyways so that we can compare.


i just used a couple of equations. 3 cubics and one linear. Sometimes these simple a level questions get to me. i was putting the sum into X and somehow trying some bull****. After the equations i just factorised and simplified and divided through by the root in the end.


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Original post by atsruser
For a standard quadratic, you have α+β=−ba,αβ=ca\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a} where α,β\alpha, \beta are the roots.

There is a similar, but more involved, result that you can look up for the standard cubic. The result follows fairly easily once you know that result and you use the fact given in the question.


yeh mate, thats a proper neat way of doin it. I'm going to use that now.


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Original post by physicsmaths
yeh mate, thats a proper neat way of doin it. I'm going to use that now.


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I personally would say that's the only 'proper' way of doing it. I say proper because I'm sure that's what the question-setter was after.
Good to see some nice hard a level questions though. The a levels nowadays are way to easy(im not complaining).


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Original post by DomStaff
I personally would say that's the only 'proper' way of doing it. I say proper because I'm sure that's what the question-setter was after.


dw my methods are normally always in the (alternative methods) section.


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ill post some other 1968 questions if anyones interested. This was probably the hardest though.


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Original post by physicsmaths
ill post some other 1968 questions if anyones interested. This was probably the hardest though.


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Sure.
Original post by physicsmaths
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Original post by Phoebe Buffay
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Original post by atsruser
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Original post by DomStaff
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After comparing the coefficients of the equality below, did y'all end up with a=1?a=1 ? That's what I got in my final expression after ending up with b3=4(bc−2d)b^3 = 4(bc - 2d), so I just said that it's equal to the form given. I don't think that's right though. Shouldn't it be more general than a=1?a=1 ?

ax3+bx2+cx+d=(x−α1)(x−α2)(x−α3)⏟∵ coeff of x3=1, a=1ax^3 + bx^2 + cx + d = \underbrace{(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)}_{\because \text{ coeff of } x^3 = 1, \ a=1}
(edited 9 years ago)
Original post by Khallil
After comparing the coefficients of the equality below, did y'all end up with a=1?a=1 ? That's what I got in my final expression after ending up with b3=4(bc−2d)b^3 = 4(bc - 2d), so I just said that it's equal to the form given. I don't think that's right though. Shouldn't it be more general than a=1?a=1 ?


Not necessarily.

Consider (2x-3)(2x-5)(x-4), expand it, and the equation still holds.

I am just about to start my proof to this.
Original post by DomStaff
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I thought so. I'll try another method. :smile:

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