The Student Room Group

Roots of polynomials

Screenshot 2022-04-25 at 22.19.44.pngwhy has their method only yielded one solution whereas mine has two distinct solutionsScreenshot 2022-04-25 at 22.20.18.pngIMG_2759599C5C76-1.jpeg
(edited 1 year ago)
If you're talking about how they seem to only get the solutions k=2 and a=1 literally read the last few lines of the solution, "k^2 = 4 -> k=+-2" and "Note if k = -2, then alpha = 7, which would yield the same roots."
Original post by MouldyVinegar
If you're talking about how they seem to only get the solutions k=2 and a=1 literally read the last few lines of the solution, "k^2 = 4 -> k=+-2" and "Note if k = -2, then alpha = 7, which would yield the same roots."

No not my point ,I said that this ms has only one distinct solution in the roots regardless of alpha and k since as you said they yield the same roots hence why I grouped non-distinct solutions up in my working out. My question is asking if my other solution is valid or not?
Original post by Student 999
No not my point ,I said that this ms has only one distinct solution in the roots regardless of alpha and k since as you said they yield the same roots hence why I grouped non-distinct solutions up in my working out. My question is asking if my other solution is valid or not?

Your extra solutions dont match the quadratic and linear terms.

Note both workings could have been considerably simplified if you'd (and the model solution) centered the roots, so
(m-3d), (m-d), (m+d), (m+3d)
where m is the mean (midpoint) and 2d is k. So using the cubic term (sum of roots)
4m = 16, m=4
and using your constant term you could write down
(16-d^2)(16-9d^2) = 105
Youd still have 2 extraneous solutions, but it would be a simple hidden quadratic. If you went with the quadratic term youd get
6*16 - 10d^2 = 86
so d=+/-1, and k is double that.

Even if you did use the constant term, as in your solution, it would have been better to solve for k as a simple symmetry argument would tell you you'd get a hidden quadratic in k, rather than a full blown quartic in alpha. However, just thinking about the order of the equations youre setting up should tell you its better to use the cubic and quadratic terms as you'd be solving linear and quadratic equations, rather than cubics and quartics. You're almost doomed to find extra solutions (extraneous) using higher powers.

So using a centered sequence with the cubic and quadratic terms (linear and quadratic equations) would be the way to go. For 7 marks your working seems way OTT and you'd still have to verify the (extraneous or not) solutions.

Edit - you might want to think about why when you match the constant term by multiplying the 4 roots, you're "guaranteed" to get an extra solution like this.
(edited 1 year ago)
Original post by mqb2766
Your extra solutions dont match the quadratic and linear terms.

Note both workings could have been considerably simplified if you'd (and the model solution) centered the roots, so
(m-3d), (m-d), (m+d), (m+3d)
where m is the mean (midpoint) and 2d is k. So using the cubic term (sum of roots)
4m = 16, m=4
and using your constant term you could write down
(16-d^2)(16-9d^2) = 105
Youd still have 2 extraneous solutions, but it would be a simple hidden quadratic. If you went with the quadratic term youd get
6*16 - 10d^2 = 86
so d=+/-1, and k is double that.

Even if you did use the constant term, as in your solution, it would have been better to solve for k as a simple symmetry argument would tell you you'd get a hidden quadratic in k, rather than a full blown quartic in alpha. However, just thinking about the order of the equations youre setting up should tell you its better to use the cubic and quadratic terms as you'd be solving linear and quadratic equations, rather than cubics and quartics. You're almost doomed to find extra solutions (extraneous) using higher powers.

So using a centered sequence with the cubic and quadratic terms (linear and quadratic equations) would be the way to go. For 7 marks your working seems way OTT and you'd still have to verify the (extraneous or not) solutions.

Edit - you might want to think about why when you match the constant term by multiplying the 4 roots, you're "guaranteed" to get an extra solution like this.



Thanks for the advice had forgotten about being able to centre around the roots. The only thing I could think of in getting extraneous solutions in the constant term is from certain roots such as alpha taking the form -alpha and so on since its a quartic having two of the roots taking negatives so
-a , a + k , -a-2k ,a+3k resulting in values of k and alpha that doesn't satisfy the initial roots
Original post by Student 999
Thanks for the advice had forgotten about being able to centre around the roots. The only thing I could think of in getting extraneous solutions in the constant term is from certain roots such as alpha taking the form -alpha and so on since its a quartic having two of the roots taking negatives so
-a , a + k , -a-2k ,a+3k resulting in values of k and alpha that doesn't satisfy the initial roots

If you fit the cubic and quadratic terms, then it should be fairly clear that there is a single sequence. The cubic term gives you the mean of the roots and the quadratic term will give two solutions for k or d, but these are +/- so generate the same sequence (four roots)

If you fit the constant term, so the product of the roots, then obviously two of the four solutions for k or d will be the correct ones. To see why you get another pair, think about when k~0. Then the product of the roots is ~256, which is larger than the constant term. As k increases, the smallest root will head towards zero so the product will reduce until its zero when the minimum root is zero. At some point, you'll have passed the value which gives the constant term. In this case, thats the correct solution. However, if you keep increasing k, the minimun root will be negative and so the product will be negative until the second smallest root hits zero. Keep increasing k, then the product will become postive again and start increasing until it its the constant term which gives the second solution for k with two negative and two positive roots. In this case, its an extraneous solution (and its mirror). So for the scenario given above youll always have one pair of extraneous solutions fitting the constant term.
(edited 1 year ago)
Original post by mqb2766
If you fit the cubic and quadratic terms, then it should be fairly clear that there is a single sequence. The cubic term gives you the mean of the roots and the quadratic term will give two solutions for k or d, but these are +/- so generate the same sequence (four roots)

If you fit the constant term, so the product of the roots, then obviously two of the four solutions for k or d will be the correct ones. To see why you get another pair, think about when k~0. Then the product of the roots is ~256, which is larger than the constant term. As k increases, the smallest root will head towards zero so the product will reduce until its zero when the minimum root is zero. At some point, you'll have passed the value which gives the constant term. In this case, thats the correct solution. However, if you keep increasing k, the minimun root will be negative and so the product will be negative until the second smallest root hits zero. Keep increasing k, then the product will become postive again and start increasing until it its the constant term which gives the second solution for k with two negative and two positive roots. In this case, its an extraneous solution (and its mirror). So for the scenario given above youll always have one pair of extraneous solutions fitting the constant term.

Thanks haven't considered the idea of how k changes with roots but after a few reads it's become a bit clearer. If you don't mind me asking another question is that when they multiplied by another quadratic term to rationalise and remove the sqrt hasn't that alone introduced more solutions making it a quartic? Or is it because when making the substitution initially they considered the positive set of solutions for x so by multiplying by the negative set you'll obtain all solutions for the question Screenshot 2022-04-26 at 19.31.27.jpgScreenshot 2022-04-26 at 19.31.53.png
Original post by Student 999
? Or is it because when making the substitution initially they considered the positive set of solutions for x so by multiplying by the negative set you'll obtain all solutions for the question

This. Its one of those questions where you could almost blindly churn out the algebra and arrive at the solution, and thinking about it is almost harder. But you could abuse notation a bit by following their working with +/-sqrt(w-1) to
w^2 + 1 = +/- 5sqrt(w-1)
and square both sides to get their result. This combines both sets of solutions from
w^2 + 5sqrt(w-1) + 1 = 0
w^2 - 5sqrt(w-1) + 1 = 0
Into a single quartic and is pretty much what you do (reverse) when you get the quadratic formula by completing the square. Its probably more obvious than a seemingly arbitrary multiply by w^2-5sqrt()+1, and preserves the roots so that you don't throw away two solutions then arbitrarily reintroduce them later on which they don't comment on / I dont like.

Its worth noting that two of the roots are complex (conjugate). Not sure if they are aware of this. Doesnt change things, but ...

Will update a bit later, but the above is enough to answer the question.
(edited 1 year ago)

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