SOLUTION: 3. if one root of a function is 1/2 - 2i, what is the equation of this function? (a) 4x^2 - 17x + 4 = 0 (b) 4x^2 - 4x + 17 = 0 (c) x^2

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Question 174621This question is from textbook amscos preparing for the regents examination mathematics b
: 3. if one root of a function is 1/2 - 2i, what is the equation of this function?
(a) 4x^2 - 17x + 4 = 0
(b) 4x^2 - 4x + 17 = 0
(c) x^2 - x + 17 = 0
(d) 4x^2 + 4x + 17 = 0
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This question is from textbook amscos preparing for the regents examination mathematics b

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
if one root of a function is 1/2 - 2i, what is the equation of this function?
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If the coefficients are Real Numbers then 1/2 + 2i must be a root.
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EQUATION:
(x - (1/2 - 2i)(x - (1/2 + 2i)) = 0
Rearrange:
((x - 1/2) + 2i)((x - 1/2) - 2i) = 0
This is the form (a+b)(a-b) which = a^2 - b^2
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(x-1/2)^2 - (2i)^2 = 0
x^2 -x + (1/4) - 4i^2 = 0
x^2 -x + 17/4 = 0

4x^2 - 4x + 17 = 0
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Ans: b
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Cheers,
Stan H.
(a) 4x^2 - 17x + 4 = 0
(b) 4x^2 - 4x + 17 = 0
(c) x^2 - x + 17 = 0
(d) 4x^2 + 4x + 17 = 0

SOLUTION: 3. if one root of a function is 1/2 - 2i, what is the equation of this function? (a) 4x^2 - 17x + 4 = 0 (b) 4x^2 - 4x + 17 = 0 (c) x^2 - x + 17 = 0 (d) 4x^2 + 4