Question 1017157: If 4 + I and 4 - I are roots of the equation z^2 + az + b = 0 find the value of a and the value of b?
For the format: ax² + bx + c = 0
Sum of the roots = - b/a
Product of the roots = c/a
given: (4 + i) and (4 - i)
8 = - b/a
17 = c/a
a = (- b)/8 = 1
- b = 8
b = - 8
a = c/17 = 1
c = 17
equation is: z² - 8z + 17 = 0
check...do something
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52767)   (Show Source):
You can put this solution on YOUR website! .
If 4 + I and 4 - I are roots of the equation z^2 + az + b = 0 find the value of a and the value of b?
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Since the quadratic equation has the coefficient "1" as the leading coefficient (at  ), then
z1 + z2 = -a,
Z1*z2 = b.
Hence, a = (4+i)*(4-i) = 16 -  = 16 - (-1) = 16 + 1 = 17.
b = (4+i) + (4-i) = 8.
That is all.
Answer by MathTherapy(10549)  (Show Source):
You can put this solution on YOUR website! If 4 + I and 4 - I are roots of the equation z^2 + az + b = 0 find the value of a and the value of b?
For the format: ax² + bx + c = 0
Sum of the roots = - b/a
Product of the roots = c/a
given: (4 + i) and (4 - i)
8 = - b/a
17 = c/a
a = (- b)/8 = 1
- b = 8
b = - 8
a = c/17 = 1
c = 17
equation is: z² - 8z + 17 = 0
check...do something
Who writes these RIDICULOUS math problems?
The general form of a quadratic equation is:  , or in this case:  . Why give:  ?
Why would a quadratic equation be written with the NORMAL variable on  : "a" as the variable on "x", and the NORMAL
variable on x: "b," as the constant: "c."? Why do these people make these RIDICULOUSLY confusing math problems?
Anyway, you're correct, the variable on z, or  , and the constant, or variable,  , for the quadratic equation: 
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