The roots of the equation x^4  3x^2 + 5x  2 = 0 are a, b, c, d. By the relation y=x^2, or otherwise, show that a^2, b^2, c^2, d^2 are the roots of the equation y^4  6y^3 + 5y^2  13y + 4 = 0. The 2nd part of the question asks: find the value of s~2 where s~n = a^n + b^n + c^n + d^n. and hence show that s~8 = 6(s~6)  5(s~4) + 62 .
How many cups equal a pint
breana. When you want to post a question, don't piggy back on another post. Click the "post a new question" at the top of the page and post what you wish. That way you will get the computer to alert you to the answer.
There are two cups to a pint (U. S. measures).
I answer part 1 below for Lee. Check that.
I'm interested to know where you're studying this algebra? I would be really impressed if this were in high school, but I'm very suspicious.
i'm studying this in secondary school advanced math class
Well now I'm impressed. This would be covered in a modern algebra course usually at the sophomore level in college.
I'm curious to know what's been covered so far. I think the result for the second part uses Newton's recursive formula for the sum of nth powers. Is that something you've covered?
no, unfortunately i have not covered that. i do not go to school in the states, this may be why the math levels are different.
Possibly, because this would be college material here. I've worked these kinds of problems, but not recently. They aren't too difficult, just tedious.
For the 2nd part I'm not sure what sn is, so I really don't know how the relation is supposed to proceed. I understand that the RHS is the sum of nth powers, but I don't get the sn on the LHS. Is there anything else in your text, some notation we're supposed to know?
well s~n is like sigma to the nth power (if that helps). no i can't find a similar question in my text and this question is as stands
Ok Jack, I think I understand the notation.
The LHS sn means s_n the sum of the nth powers of the roots.
Here's what is asked:
(1) s2=a^2 + b^2 + c^2 + d^2
(2) s4=a^4 + b^4 + c^4 + d^4
(3) s6=a^6 + b^6 + c^6 + d^6
(4) s6=a^8 + b^8 + c^8 + d^8
We want to show
(4)=6(3)5(2)+62
I don't see how to get an 8th deg poly = to some linear combination of lower degree polys.
i understand what you are saying but my problem is simplifying it when i use that method. i cannot think of anyway else to prove it though. much thanks anyways.
Sure. The only thing I can add is that from the original equation in part 1 we have sigmas 1,2,3,and4 To calculate any higher power of sn we need the sigmas up to that n. I'm not sure how s6 and s8 are supposed to be done here. I know there are other sites you can check that specialize in higher math. Just google homework help math. I've posted on a couple of them, but sometimes it can take a day or two to get a respone.
Ok, I made some progress in understanding the notation and I can report that there's a lot of tedious calculating to do. If we have
x^4+px^3+qx^2+rx+s=0
p=sigma1, q=sigma2, r=sigma3, s=sigma4
s1=p
s2=p^22q
s3=(p)^3 3(p)q + 3(r)
s4=(p^22q)^2  2(q^2  2(p)(r))
s6=(s3)^2  2qs
Verify s6 is correct, I didn't carry out all the multiplications.
After you make the substitutions with the coefficients from the first part you should be able to find s8.
After reviewing the problem I don't think there's as much calculating as I thought.
The coefficients of the first equation are (1,0,3,5,2)
If you negate the coefficients of the odd degree terms and multiply them together you have
(1,0,3,5,2)(1,0,3,5,2)=(1,0,6,0,5,0,13,0,4)
Keep in mind that these are the coefficients, so you should construct some type of diagram to see how they're caluculated. The coefficients on the RHS are those of the second polynomial.
You should see that the second nonzero coeffic. is s2.
If you repeat this proces using
(1,6,5,13,4)(1,6,5,13,4)=??
Then then second nonzero coeffic is s4
You'll need to repeat this two more time to get s6 and s8
No matter how you do this there will be some calculating in order to get the sum of the 8th power of the roots. I'd be glad to check your work.
Ok, after checking what I posted I see the process only needs to be repeated once more to get s8, not s6. You'll need to compute s6 separately.
I found s2=6,s4=26,s6=165 and s8=922 and the equation you were asked to check holds:
922=6*1655*26+62
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