Quadratic Equation with imaginary roots. I know that if the discriminant of a quadratic equation is less than $0$, the roots are imaginary.
But why is this quadratic expression (with imaginary roots) always positive for all values of $x$?
Can you explain me the logic? My text book has directly stated that fact. 
Thanks.
 A: Recall the geometric interpretation for the quadratic equation
$$ax^2+bx+c=0$$
which is the solution of the system


*

*$y=ax^2+bx+c$

*$y=0$
which represents the intersection of a parabola with the $x$ axis and we can have three cases


*

*$2$ real solutions that is the parabola intersects the $x$ axis ($\Delta >0$)

*$1$ real solution that is the parabola is tangent to $x$ axis ($\Delta =0$)

*$2$ complex solutions that is the parabola does not intersect the $x$ axis ($\Delta <0$)



and in the latter case the expression is positive or negative depending upon the sign of the coefficient $a$.
A: Because any such quadratic breaks into a linear polynomial's square plus positive contant. Hence for any value of x the quadratic stays +ve
Also be advised a quadratic expression with complex roots may either be +ve or -ve for all values ox depending on the sign of coefficient of $x^2$
A: Consider the quadratic polynomial
$$
a_2x^2+a_1x+a_0.
$$
If depends on the sign of $a_2$ if the polynomial has to be always positive or always negative.
If $a_2>0$ then the quadratic polynomial is positive for large $x$. 
Assume that the quadratic polynomial has somewhere a negative value.
Then by Mean Value Theorem, you will find two real zeros. Since a quadratic polynomial has at most two zeros, it can't have further imaginary roots.
If $a_2<0$ with the same argument, you get that either the quadratic polynomial has imaginary roots or it is always negative.
A: It doesn't have to be.
It's possible that the quadratic is always negative for all real $x$.
But what is true is that if the quadratic has only complex roots (has not real roots) then it must be either positive for all real $x$ or negative for all real $x$.
Why?
Well, if it were positive for some real $x_1$ and it were negative for some real $x_2$, then it'd have to be zero for some real $c$ in between $x_1$ and $x_2$.  That that would mean $c$ is a real root.
(Furthermore, you can tell if it will always be positive or negative by checking if coeeficient of $x^2$  ... or the constant term.)
