Question 961545: The question is to find all quadratic equations with roots 2+3i and 2-3i. I have actually been given an answer by my teacher, but I wish to find out the logic behind this. My teacher has given me a solution, which is found below.
sum=4=-b/a
Product=4+9=13=c/a
=> a=1, b=-4, c=13
a(x^2-4x+13)=0
So, there it is. I understand how to perform sums like this with square roots and normal integers in the place of i, but I do not understand where to even start with i in the equation, as I have tried the usual method that I use for numbers and roots and the answer I received was incorrect. Any help would be much appreciated.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The question is to find all quadratic equations with roots 2+3i and 2-3i. I have actually been given an answer by my teacher, but I wish to find out the logic behind this. My teacher has given me a solution, which is found below.
sum=4=-b/a
Product=4+9=13=c/a
=> a=1, b=-4, c=13
a(x^2-4x+13)=0
So, there it is. I understand how to perform sums like this with square roots and normal integers in the place of i, but I do not understand where to even start with i in the equation, as I have tried the usual method that I use for numbers and roots and the answer I received was incorrect. Any help would be much appreciated.
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Form:: y = ax^2+bx+c
Roots:: x = [-b+sqrt(b^2-4ac)]/(2a) and x = [-b-sqrt(b^2-4ac)]/(2a)
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Adding them you get: 2(-b/(2a)) = -b/a
Multiplying them you get: (-b/(2a))^2-(sqrt(b^2-4ac))^2/(2a)^2)
= (b^2/4a^2 - (b^2-4ac))/(4a^2)
= (b^2 - b^2 + 4ac)/(4a^2) = 4ac/(4a^2) = c/a
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Your roots are 2+3i and 2-3i
Sum = 4, so -b/a = 4 or -b = 4a
Product = 4+9 = 13, So c/a = 13 or c = 13a
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Equation:
y = ax^2 -4ax + 13a
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y = a(x^2 - 4x + 13)
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Cheers,
Stan H.
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