How do we know that these roots are non-real?

Question

The roots, x and y, of a quadratic equation $$ax^2 + bx + c=0$$ satisfy the following inequality:

$$x^2 + y^2 < 0$$

What can you conclude about the nature of the roots of the quadratic equation?

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$$x^2 + y^2 = (x+y)^2 - 2xy = \frac{b^2 - 2ac}{a^2}$$

since $$x^2 + y^2 <0$$, we can say $$b^2 - 2ac < a^2$$ therefore, $$b^2 - 4ac < a(a-2c)$$ therefore,$$ D < a(a-2c)$$

How do I proceed from here?

Answer 1

Alternatively: $$x^2+y^2=\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)^2+\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)^2=\frac{b^2-4ac}{a^2}<0 \Rightarrow$$ $$b^2-4ac<0 \Rightarrow x,y\in \mathbb{C}.$$

Answer 2

Since $x^2 + y^2<0$, they can't be real because squares of real numbers are non-negative.