SOLUTION: Somehow my jamielee90@ymail.com address doesnt seem to be working since I never seem to recieve a msg's from this site, on that account, so I asked a friend to make me an account o

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Question 199208: Somehow my jamielee90@ymail.com address doesnt seem to be working since I never seem to recieve a msg's from this site, on that account, so I asked a friend to make me an account on his, hope it goes through now =/
Ok so to the real problem, I dont get how to solve these problems, could someone help me out here?
1)Find a quadratic equation with roots (4+i) and (4-i).
2)Find the irrational roots of x^3-4x^2+2x+1=0 (I should also use the quadratic formula to solve it)
3) x^3-2x^2+x-3=0 what are the possible rational roots?
As you might see I am not too good with roots, I never really have been =(
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x=4%2Bi or x=4-i Start with the given solutions.


x-4=i or x-4=-i Subtract 4 from both sides (for each equation).


%28x-4%29%5E2=i%5E2 or %28x-4%29%5E2=%28-i%29%5E2 Square both sides


%28x-4%29%5E2=-1 or %28x-4%29%5E2=-1 Square i to get -1 and square -i to get -1


%28x-4%29%5E2%2B1=0 or %28x-4%29%5E2%2B1=0 Add 1 to both sides.


Since the equations are the same, we can focus on one equation:


%28x-4%29%5E2%2B1=0


x%5E2-8x%2B16%2B1=0 FOIL


x%5E2-8x%2B17=0 Combine like terms.


So the quadratic with the roots 4+i and 4-i is y=x%5E2-8x%2B17




# 2


First, let's find the possible rational zeros


Any rational zero can be found through this equation

where p and q are the factors of the last and first coefficients


So let's list the factors of 1 (the last coefficient):



Now let's list the factors of 1 (the first coefficient):



Now let's divide each factor of the last coefficient by each factor of the first coefficient









Now simplify

These are all the distinct rational zeros of the function that could occur




Let's see if the possible zero 1 is really a root for the function x%5E3-4x%5E2%2B2x%2B1


So let's make the synthetic division table for the function x%5E3-4x%5E2%2B2x%2B1 given the possible zero 1:
1|1-421
| 1-3-1
1-3-10

Since the remainder 0 (the right most entry in the last row) is equal to zero, this means that 1 is a zero of x%5E3-4x%5E2%2B2x%2B1

Take note that the first three values in the bottom row are 1, -3, and -1. So this means that


x%5E3-4x%5E2%2B2x%2B1=%28x-1%29%28x%5E2-3x-1%29


Now all we need to do is solve x%5E2-3x-1=0 to find the next two zeros:


Notice we have a quadratic in the form of Ax%5E2%2BBx%2BC where A=1, B=-3, and C=-1


Let's use the quadratic formula to solve for "x":


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%28-3%29+%2B-+sqrt%28+%28-3%29%5E2-4%281%29%28-1%29+%29%29%2F%282%281%29%29 Plug in A=1, B=-3, and C=-1


x+=+%283+%2B-+sqrt%28+%28-3%29%5E2-4%281%29%28-1%29+%29%29%2F%282%281%29%29 Negate -3 to get 3.


x+=+%283+%2B-+sqrt%28+9-4%281%29%28-1%29+%29%29%2F%282%281%29%29 Square -3 to get 9.


x+=+%283+%2B-+sqrt%28+9--4+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-1%29 to get -4


x+=+%283+%2B-+sqrt%28+9%2B4+%29%29%2F%282%281%29%29 Rewrite sqrt%289--4%29 as sqrt%289%2B4%29


x+=+%283+%2B-+sqrt%28+13+%29%29%2F%282%281%29%29 Add 9 to 4 to get 13


x+=+%283+%2B-+sqrt%28+13+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%283%2Bsqrt%2813%29%29%2F%282%29 or x+=+%283-sqrt%2813%29%29%2F%282%29 Break up the expression.


So the next two zeros are x+=+%283%2Bsqrt%2813%29%29%2F%282%29 or x+=+%283-sqrt%2813%29%29%2F%282%29


===================================================================================

Answer:

So the three roots are x=1, x+=+%283%2Bsqrt%2813%29%29%2F%282%29 or x+=+%283-sqrt%2813%29%29%2F%282%29


where the irrational roots are x+=+%283%2Bsqrt%2813%29%29%2F%282%29 and x+=+%283-sqrt%2813%29%29%2F%282%29






# 3

Any rational zero can be found through this equation

where p and q are the factors of the last and first coefficients


So let's list the factors of -3 (the last coefficient):



Now let's list the factors of 1 (the first coefficient):



Now let's divide each factor of the last coefficient by each factor of the first coefficient









Now simplify

These are all the distinct rational zeros of the function that could occur