SOLUTION: Could someone help me with these, I had it last year but forgot how it works: 1) Find a quadratic equation with roots (4+i)and(4-i). 2) Given the equation x^3-2x^2+x-3=0, what

Algebra ->  Square-cubic-other-roots -> SOLUTION: Could someone help me with these, I had it last year but forgot how it works: 1) Find a quadratic equation with roots (4+i)and(4-i). 2) Given the equation x^3-2x^2+x-3=0, what       Log On


   



Question 198572: Could someone help me with these, I had it last year but forgot how it works:
1) Find a quadratic equation with roots (4+i)and(4-i).
2) Given the equation x^3-2x^2+x-3=0, what are the possible rational roots??
3) Find the irratinal roots of the equation; x^3-4x^2+2x+1=0 I should use the quadratic formula to solve the equation, it might sound stupid but I never really understood how to use the quadratic formula =(

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

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






# 3


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