# How you can tell that the quadratic function y=x^2 +6x+15 has no real zeroes without graphing the function?

Nov 16, 2016

When given a quadratic equation in standard form:

$y = a {x}^{2} + b x + c$ or $x = a {y}^{2} + b y + c$

Check the value of the determinant:

${b}^{2} - 4 \left(a\right) \left(c\right)$

If it is greater than zero, the equation will have two real roots.
If it is equal to zero, the equation will have one root (a.k.a. a repeated root).
If it is less than zero, the equation will have complex conjugate pair of roots.

In the case of the given equation

${b}^{2} - 4 \left(a\right) \left(c\right) = {6}^{2} - 4 \left(1\right) \left(15\right) = - 24$

This equation will have a complex conjugate pair of roots $\left(- 3 + \sqrt{6} i\right) \mathmr{and} \left(- 3 - \sqrt{6} i\right)$ .