# [Roots of polynomials] Question Method Stuck

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#1
I’m on this question:

I understand I could solve the cubic normally but the purpose of this question is to use the deviations of the quartic roots and coefficients to solve the problem. I applied
alpha + beta + gamma + delta = -b/a
and
(alpha)(beta)(gamma)(delta) = e/a
To form two simultaneous equations.
They are horrific to work with and I think I may have used an inefficient method to solve this.

Here are my current workings:

Is there a better method to do this using the roots to coefficients link?
Any help is appreciated, thanks.
0
1 year ago
#2
You could try the relationship
alpha*beta + alpha*gamma + alpha*delta + .... = -c/a
You'd just have a quadratic in {a,k} in addition to the linear, rather than the quartic.

Edit: not worked it out though :-)
Last edited by mqb2766; 1 year ago
0
1 year ago
#3
(Original post by BrandonS15)
I’m on this question:

I understand I could solve the cubic normally but the purpose of this question is to use the deviations of the quartic roots and coefficients to solve the problem. I applied
alpha + beta + gamma + delta = -b/a
and
(alpha)(beta)(gamma)(delta) = e/a
To form two simultaneous equations.
They are horrific to work with and I think I may have used an inefficient method to solve this.

Here are my current workings:

Is there a better method to do this using the roots to coefficients link?
Any help is appreciated, thanks.

Couple of ideas:

Exploiting the AP nature of the roots, I'd let r say, equal their mean.

The sum of the roots is then simply 4r, and the product is a disguised quadratic (in k^2).

2nd.

Not using the coefficent to roots as such, but rather recognising the roots must be +/- factors of 105, and it drops out in a small number of lines.

Edit: Corrected r^2 to k^2.
Last edited by ghostwalker; 1 year ago
2
#4
(Original post by ghostwalker)
Couple of ideas:

Exploiting the AP nature of the roots, I'd let r say, equal their mean.

The sum of the roots is then simply 4r, and the product is a disguised quadratic (in k^2).

2nd.

Not using the coefficent to roots as such, but rather recognising the roots must be +/- factors of 105, and it drops out in a small number of lines.

Edit: Corrected r^2 to k^2.
Your first method, I’m not sure I fully understand how to incorporate it into the given question. I understand the roots are AP but I don’t understand how the mean of the roots can be used and by product do you mean:
summation of the product of root pairs?
(alpha)(beta) + (alpha)(gamma) + (alpha)(delta) ...?
Is this where the quadratic is coming from because wouldn’t it have two missing variables - alpha and k and to use r in terms of alpha and k I’d end up reversing the previous mean manipulation and solve the linear and quadratic in terms of alpha and k which I currently did based on mqb2766 method.
0
1 year ago
#5
One method is noting that:
and
are solutions.

Then you can divide the quatic formula by:

The result is a quadratic which then can be easily factorised.

Otherwise as and are solutions and is not, we can see if and by testing if is a solution and so on.
Last edited by simon0; 1 year ago
0
1 year ago
#6
(Original post by BrandonS15)
Your first method, I’m not sure I fully understand how to incorporate it into the given question. I understand the roots are AP but I don’t understand how the mean of the roots can be used and by product do you mean:
summation of the product of root pairs?
(alpha)(beta) + (alpha)(gamma) + (alpha)(delta) ...?
Is this where the quadratic is coming from because wouldn’t it have two missing variables - alpha and k and to use r in terms of alpha and k I’d end up reversing the previous mean manipulation and solve the linear and quadratic in terms of alpha and k which I currently did based on mqb2766 method.
For the mean part, note the mean of the 4 roots is
a + 3k/2
When you divide your b/a equation by 4, you get exactly this
a + 3k/2 = -b/4
(The "a" in the quartic is 1 and not the a on the left hand side, so I've not included it on the right).
The mean of the roots is always -b/4 (for a quartic with unity x^4 term), but its a good spot to work with a centered AP sequence in the first place.

When you then work with the "-e" equation (product of four roots) you have a product of two DOTS (difference of two squares) when the roots are expresed as
mean +/- k/2 or 3k/2
because of the symmetry. Anything with DOTS in is usually the elegant way to solve the problem so it was a good spot by ghostwalker and if you've done it using the -c (sum of products of two roots) way I proposed, have a go this way as well. It will cut down the algebra and make mistakes less likely.
Last edited by mqb2766; 1 year ago
0
1 year ago
#7
(Original post by BrandonS15)
Your first method, I’m not sure I fully understand how to incorporate it into the given question. I understand the roots are AP but I don’t understand how the mean of the roots can be used and by product do you mean:
summation of the product of root pairs?
(alpha)(beta) + (alpha)(gamma) + (alpha)(delta) ...?
Is this where the quadratic is coming from because wouldn’t it have two missing variables - alpha and k and to use r in terms of alpha and k I’d end up reversing the previous mean manipulation and solve the linear and quadratic in terms of alpha and k which I currently did based on mqb2766 method.
mqb2766 has covered most of it. So, I'll just expand a few points.

If r is the mean, then the terms in your AP (roots), become r +/- multiples of k/2.

To simplify the algebra, let s= k/2.

Then our roots are r-3s, r-s, r+s, r+3s.

We know what r is, 4.

If we now work with the product of all 4 roots, equals 105.

We end up with a quadratic in s^2. Solving s=+/-1, or +/- root(151)/3

We can ignore the second set of possibiities since the roots must be rational (and r+s wouldn't be with those latter values).

And it doesn't matter whether we choose s=1 or s=-1 due to the symmetry of the four AP terms about the mean.
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