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.

• We do you say it is always positive? It could be always negative. But if it were positive for one $x_1$ and negative for an $x_2$ then it must be $0$ for some $x$ value in between. – fleablood Jun 16 '18 at 19:43
• Think geometrically: if $p(x) = ax^2 + bx + c$ has no real roots, that means that its graph does not intersect the $x$-axis. Since the graph of $p$ is a parabola, there are two possibilities: either the parabola opens upward and lies entirely above the $x$-axis (this occurs when $a > 0$, here $p(x) > 0$ for all $x$), or the parabola opens downward and lies entirely below the $x$-axis (this occurs when $a < 0$, here $p(x) < 0$ for all $x$). – Xander Henderson Jun 18 '18 at 2:03
• @William Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – user Aug 4 '18 at 22:11
• @gimusi I'm sorry, I didn't know that. I'm new :) I'll accept your answer as it seems to be the best. – William Aug 5 '18 at 5:06
• @William You are welcome! Thanks – user Aug 5 '18 at 6:43

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 $a$.

• Thnx for the clear explanation :-) – William Jun 18 '18 at 6:23
• You are welcome! Bye – user Jun 18 '18 at 6:41

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 $x^2$

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.

• What ...is... you talking about lol? :o!! – William Jun 16 '18 at 19:22

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.)