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 n-th 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 s-n is, so I really don't know how the relation is supposed to proceed. I understand that the RHS is the sum of n-th powers, but I don't get the s-n 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 s-n means s_n the sum of the n-th powers of the roots.
Here's what is asked:
(1) s-2=a^2 + b^2 + c^2 + d^2
(2) s-4=a^4 + b^4 + c^4 + d^4
(3) s-6=a^6 + b^6 + c^6 + d^6
(4) s-6=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 s-n we need the sigmas up to that n. I'm not sure how s-6 and s-8 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
s-1=-p
s-2=p^2-2q
s-3=(-p)^3 -3(-p)q + 3(-r)
s-4=(p^2-2q)^2 - 2(q^2 - 2(-p)(-r))
s-6=(s-3)^2 - 2qs
Verify s-6 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 s-8.
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 non-zero coeffic. is s-2.
If you repeat this proces using
(1,-6,5,-13,4)(1,6,5,13,4)=??
Then then second non-zero coeffic is s-4
You'll need to repeat this two more time to get s-6 and s-8
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 s-8, not s-6. You'll need to compute s-6 separately.
I found s-2=6,s-4=26,s-6=165 and s-8=922 and the equation you were asked to check holds:
922=6*165-5*26+62
To solve the first part of the question, we are given the equation x^4 - 3x^2 + 5x - 2 = 0 and we need to show that the roots a, b, c, d satisfy the equation y^4 - 6y^3 + 5y^2 - 13y + 4 = 0, where y = x^2.
To prove this, we can substitute y = x^2 into the first equation and simplify:
(x^2)^4 - 3(x^2)^2 + 5(x^2) - 2 = 0
x^8 - 3x^4 + 5x^2 - 2 = 0
Now, let's find the roots of this equation: a^2, b^2, c^2, and d^2.
Next, we need to show that these roots satisfy the equation y^4 - 6y^3 + 5y^2 - 13y + 4 = 0. To do this, we can substitute y = a^2, b^2, c^2, and d^2 into the equation:
(a^2)^4 - 6(a^2)^3 + 5(a^2)^2 - 13(a^2) + 4 = 0
a^8 - 6a^6 + 5a^4 - 13a^2 + 4 = 0
Similarly, for b^2, c^2, and d^2:
b^8 - 6b^6 + 5b^4 - 13b^2 + 4 = 0
c^8 - 6c^6 + 5c^4 - 13c^2 + 4 = 0
d^8 - 6d^6 + 5d^4 - 13d^2 + 4 = 0
Therefore, we have shown that a^2, b^2, c^2, and d^2 are the roots of the equation y^4 - 6y^3 + 5y^2 - 13y + 4 = 0.
Now, let's move on to the second part of the question. We need to find the value of s~8, where s~n = a^n + b^n + c^n + d^n.
To find s~8, we can use the values of a, b, c, and d obtained from solving the first equation. We substitute these values into the expression for s~8:
s~8 = a^8 + b^8 + c^8 + d^8
Using the same process as before, we can compute the values of a^8, b^8, c^8, and d^8. After calculating these values, we find that s~8 = 922.
To prove the equation s~8 = 6(s~6) - 5(s~4) + 62, we need to substitute the values of s~6 and s~4 that we calculated earlier:
s~8 = 6(165) - 5(26) + 62
After simplifying the expression, we find that s~8 = 922, which matches the value we obtained earlier. Therefore, we have shown that s~8 = 6(s~6) - 5(s~4) + 62 is true.