Quadratic Equation with imaginary roots.

Question

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.

Answer 1

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 $a$.

Answer 2

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 $x^2$

Answer 3

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.

Answer 4

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.)