help on this one.. working out shown (1 Viewer)
- Thread starter bos1234
- Start date
SoulSearcher
Active Member
- Joined
- Oct 13, 2005
- Messages
- 6,757
- Gender
- Male
- HSC
- 2007
ax3 + bx2 + cx + d = 0
Sum of roots of the above polynomial (roots denoted as x, y and z)
x + y + z = -b/a
You can see it from there.
Sum of roots of the above polynomial (roots denoted as x, y and z)
x + y + z = -b/a
You can see it from there.
missAy
New Member
What's the difference between show and prove? 
idling fire
Member
I don't think it would matter if you used the 'normal' way of showing it. After all, it counts as a proof.
And I wouldn't have a clue as to how to do it that way. ^^
Just from theory, the equation becomes
P(x)= x^3 - (sum of roots)x^2 + (sum of roots multiplied in groups of 2)x - (product of roots)
It's from the theory, and as far as I know, doesn't require such proof...
And I wouldn't have a clue as to how to do it that way. ^^
Just from theory, the equation becomes
P(x)= x^3 - (sum of roots)x^2 + (sum of roots multiplied in groups of 2)x - (product of roots)
It's from the theory, and as far as I know, doesn't require such proof...
Last edited:
You can solve this problem in one of two ways.
First Solution
==================
You can use the fact that the sum of the roots one at a time is equal to -b/a, therefore...
a + 1/a + b= L
Second Solution
==================
This solution isn't necessary in the HSC course as markers generally accept the first solution, however, if you would like to know how to do it, this is how it goes:
Since you know that the polynomial is of degree three, then it must have three roots, all of which are given (a, 1/a, b), thus....
P(x)=(x-a)(x-1/a)(x-b) which is equivalent to x3-Lx2 +Lx -M
Using the LHS:
P(x)=(x2 - ax - x/a +1)(x-b)
=x3-ax2 -x2/a + x -bx2 -abx -bx/a -b
Rearranging:
P(x)=x3-(a+1/a+b)x2 + (1-ab-b/a)x-b
By equating the two coefficients of x2 (From the equations above), we get:
-(a+1/a+b)=-L
.: a+1/a+b = L
NOTE: This is the long way to do this, and i wouldn't recommend it under exam conditions because if you stuff up the algebra you will waste time trying to go back and fix it.
EDIT: I just read over your solution and found that you did not divide by a(alpha) before you equated the coefficients, giving you an incorrect answer. You must ensure that the coefficients of the leading terms are both the same (i.e. x3) before you equate the coefficients. Doing this in an exam will stress you out and will result in A LOT OF WASTED TIME.
First Solution
==================
You can use the fact that the sum of the roots one at a time is equal to -b/a, therefore...
a + 1/a + b= L
Second Solution
==================
This solution isn't necessary in the HSC course as markers generally accept the first solution, however, if you would like to know how to do it, this is how it goes:
Since you know that the polynomial is of degree three, then it must have three roots, all of which are given (a, 1/a, b), thus....
P(x)=(x-a)(x-1/a)(x-b) which is equivalent to x3-Lx2 +Lx -M
Using the LHS:
P(x)=(x2 - ax - x/a +1)(x-b)
=x3-ax2 -x2/a + x -bx2 -abx -bx/a -b
Rearranging:
P(x)=x3-(a+1/a+b)x2 + (1-ab-b/a)x-b
By equating the two coefficients of x2 (From the equations above), we get:
-(a+1/a+b)=-L
.: a+1/a+b = L
NOTE: This is the long way to do this, and i wouldn't recommend it under exam conditions because if you stuff up the algebra you will waste time trying to go back and fix it.
EDIT: I just read over your solution and found that you did not divide by a(alpha) before you equated the coefficients, giving you an incorrect answer. You must ensure that the coefficients of the leading terms are both the same (i.e. x3) before you equate the coefficients. Doing this in an exam will stress you out and will result in A LOT OF WASTED TIME.
Last edited: