SOLUTION: Q. If one complex zero of a quadratic equation is 2 + i, what is the quadratic equation in ax^2 + bx + c = 0 thanks, Mike

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Question 5071: Q. If one complex zero of a quadratic equation is 2 + i, what is the quadratic equation in ax^2 + bx + c = 0
thanks, Mike
Found 2 solutions by Earlsdon, ichudov:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Starting with the quadratic formula:
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29 and given that:
x=2%2B-sqrt%28-1%29 we can conclude that:
-b/2a = 2 Assuming that a is a non-zero integer, it must be 1, therefore:
-b/2 = 2
So, a = 1, and b = -4
Now we need to find c. Looking at the radical contents, we can conclude that it is equal to -4.
Why? because sqrt%28b%5E2+-+4ac%29%2F2 = -1 Substituting the values of a and b, we find:
sqrt%28%28-4%29%5E2+-+%284%29%281%29c%29%2F2%29%29 = sqrt%28-1%29
sqrt%28%2816-4c%29%2F2%29 = sqrt%28-1%29 Multiply both sides by 2.
sqrt%2816+-+4c%29 = 2sqrt%28-1%29 Square both sides.
16 - 4c = 4(-1) Subtract 16 from both sides.
-4c = -20 Divide both sides by -4
c = 5
Now we can write the quadratic equation whose roots (zeros) are (2 +/- i)
x^2 - 4x + 5 = 0

Answer by ichudov(507) About Me  (Show Source):
You can put this solution on YOUR website!
This question assumes that the numbers a, b, c are real number.
Because the roots of the equation are complex, they are of form 2+i and 2-i.
The equation is a trinomial that is a product of two binomials