Asked by Please :)
How do I find complex zeros in a an equation
if the equation doesn't have the + or - value at the end can th ecomplex zero(s) be made up?
I'm getting really confused with complex zeros well zeros in general
if the equation doesn't have the + or - value at the end can th ecomplex zero(s) be made up?
I'm getting really confused with complex zeros well zeros in general
Answers
Answered by
bobpursley
I assume you are speaking of solutions in a quadratic equation
Complex solultion will be of the form
x= a+bi
Example:
x^2+x + 3=0
x= (-2 +- sqrt (1-12))/2
x= -1 +-sqrt(-11/4)
x=-1+- 0.5 sqrt(11) i
solutions 1
x= -1+.5sqrt(11) i
x= -1-.5sqrt(11) i
the imaginary part comes out of the serd, if it has a sqrt of a negative quanity you will get a complex number. If the value in the sqrt is positive, then you have a real root, not complex.
you could of course write it as complex root, such as x= 14+0i but that seems to be a little sillly, to me.
solutions
Complex solultion will be of the form
x= a+bi
Example:
x^2+x + 3=0
x= (-2 +- sqrt (1-12))/2
x= -1 +-sqrt(-11/4)
x=-1+- 0.5 sqrt(11) i
solutions 1
x= -1+.5sqrt(11) i
x= -1-.5sqrt(11) i
the imaginary part comes out of the serd, if it has a sqrt of a negative quanity you will get a complex number. If the value in the sqrt is positive, then you have a real root, not complex.
you could of course write it as complex root, such as x= 14+0i but that seems to be a little sillly, to me.
solutions
Answered by
Steve
if you're talking about quadratics with no constant term, then they look like
y = ax^2+bx
That's just
y = x(ax+b)
and it has two real roots (0 and -b/a). No complex roots are possible.
y = ax^2+bx
That's just
y = x(ax+b)
and it has two real roots (0 and -b/a). No complex roots are possible.
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