# Complex Solutions to a Quadratic Equation

Given equation: $$x^2 + 5x + c = 0$$ For what $$c$$'s will the equation have non-real solutions?

I have try to plug in all kinds of different values for $$c$$ and then solve the equation to see if it gives me complex solution.

My question: Is there a theorem/lemma or a formula I need to state to answer this question?

• Hint: Just use the Quadratic Formula and analyze that result for the roots. – Moo Sep 16 '16 at 1:26
• Try plotting it for $c=0$, and consider what happens to the roots if $c$ increases or decreases. – Semiclassical Sep 16 '16 at 1:29

You have exactly one root if you have a perfect square.

Compete the square:

$(x^2 + 2 \frac 52 x + \frac {25}{4}) = (x+\frac 52)^2$

If $c = \frac {25}{4}$ there is one (real) root.

If $c >\frac {25}{4}$ your curve never touches the x axis. And you have no real roots. If a quadratic has no real roots, it has complex roots.

if $c > \frac {25}{4}$ the curve crosses the x axis. And it crosses on both sides of the vertex.

If you know the quadratic formula, then what does it take to get a negative under the radical?

• Great explanation. I was not thinking about square root or graphing it. Thanks! – Cesar Agama Sep 16 '16 at 6:19

Hint: This question is equivalent to finding the values of $c$ when $$b^2-4ac=5^2-4c<0$$

The discriminant formula for any quadratic is-
$$D=b^2-4ac$$, where $$a,b$$ and $$c$$ are the respective coefficients of the quadratic.

• If $$D>0$$, then the quadratic has two real and distinct roots.
• If $$D=0$$, then the quadratic has two real and equal roots.
• If $$D<0$$, then the quadratic has two complex roots.

$$a=1, b=5, c=?$$ $$D=5^2 - 4×1×c =25 - 4c$$
$$D<0$$
$$\Rightarrow 25 - 4c <0$$
$$\Rightarrow c < 25/4$$ $$_{(Answer)}$$