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# Algebra 2

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1. The sequence a(n) satisfies a(1) = 3 and

a(n) = [a(n-1)-1]/a(n-1)+1

for all n ≥ 2. Find a(100).

(I'm sorry about the way I formatted this question.

2. One of the roots of 2x^3 − 9x^2 + 13x + k = 0 is x = 2.  Find the other two roots.

Jun 17, 2024

#1
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Problem 1:

To find a(100), we first need to find a(2), a(3), a(4), and so on until we reach a(100) by applying the recursive formula given.

a(1) = 3

To find a(2):
a(2) = [a(1)-1]/a(1)+1
a(2) = [3-1]/3+1
a(2) = 2/4
a(2) = 1/2

To find a(3):

a(3) = [a(2)-1]/a(2)+1

a(3) = [1/2-1]/1/2+1

a(3) = -1/2 / 3/2

a(3) = -1/3

Continuing on in this fashion, we find:

a(4) = 1/2

a(5) = 1

a(6) = 0

It's clear that the sequence oscillates between a positive number and zero.

Therefore, for n ≥ 5, a(n) = 1 for odd n and a(n) = 0 for even n.

Since 100 is even, a(100) = 0.

Jun 17, 2024
#2
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Problem 2:

To find the other two roots, we can use polynomial division or synthetic division to divide the given polynomial by (x - 2) since x = 2 is a root.

Using synthetic division:
2|2 -9 13 k
2 -4 18
___________
2 -7 9 k

Therefore, the polynomial can be factored as:

2x^3 − 9x^2 + 13x + k = (x - 2)(2x^2 - 7x + 9)

Now, to find the other two roots, we can solve the quadratic equation 2x^2 - 7x + 9 = 0 using the quadratic formula:

x = (-(-7) ± √((-7)^2 - 4*2*9)) / 2*2

x = (7 ± √(49 - 72)) / 4

x = (7 ± √(-23)) / 4

Since the discriminant is negative, the quadratic equation has two complex roots:

x = (7 ± √23i) / 4

Therefore, the other two roots are:

x = (7 + √23i) / 4

x = (7 - √23i) / 4

Jun 17, 2024
#3
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2

2 [   2    - 9    13    k  ]

4  - 10    6

__________________

2     - 5      3   k + 6

k + 6 = 0

k = -6

The  remaining polynomial  is

2x^2  - 5x  + 3  =  0

(2x - 3) ( x - 1)  = 0

The other two roots   are  3/2  and 1

Jun 17, 2024