# How do you find the discriminant and how many and what type of solutions does 4x^2+ 9 = 0 have?

Jun 15, 2018

$\Delta = {b}^{2} - 4 a c = - 144 < 0$ and hence this quadratic has a complex conjugate pair of non-Real roots.

#### Explanation:

Given:

$4 {x}^{2} + 9 = 0$

Note that we can also write this as:

$4 {x}^{2} + 0 x + 9 = 0$

which is in the standard form:

$a {x}^{2} + b x + c = 0$

with $a = 4$, $b = 0$ and $c = 9$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(\textcolor{b l u e}{0}\right)}^{2} - 4 \left(\textcolor{b l u e}{4}\right) \left(\textcolor{b l u e}{9}\right) = 0 - 144 = - 144$

Since $\Delta < 0$ we can tell that this quadratic equation has no Real roots. It has a complex conjugate pair of non-Real roots.

We can use the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

with $A = 2 x$ and $B = 3 i$

where $i$ is the imaginary unit satisfying ${i}^{2} = - 1$ to find:

$0 = 4 {x}^{2} + 9$

$\textcolor{w h i t e}{0} = {\left(2 x\right)}^{2} + {3}^{2}$

$\textcolor{w h i t e}{0} = {\left(2 x\right)}^{2} - {\left(3 i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(2 x - 3 i\right) \left(2 x + 3 i\right)$

Hence:

$x = \pm \frac{3}{2} i$