Why can't a quadratic a real and a non real roots? Explain.
Thanks
Actually it can... for example
$\displaystyle (x-2i)(x+3)=0$
if we expand it we get
$\displaystyle x^2+(3-2i)x-6i=0$
has solutions x=2i and x=-3.
The question I think you are asking is...
Why cant a quadratic with REAL coefficients have both real and complex solutions.
Every quadratic can be solved using the quadratic fomula...
The discriminant tell us what type and how many solutions we will get.
To remind you it is the part under the radical
$\displaystyle b^2-4ac$
If it is positive you will get two different real solution.
If it is zero you get one repeated real solution.
if it is negative two complex solutions.
I hope this helps...
I'm guessing you mean a Quadratic Expression with REAL coefficients. It does help to state the question clearly.
Quadratics look like this: (x+a)(x+b)
Expanded, gives: x^2 + x(a+b) + a*b
So,
a+b must be Real and a*b must be Real.
What does that say about a and b?