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

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Answer 1 · 2018-06-15T06:04:13

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

Given:

$4x^2+9 = 0$

Note that we can also write this as:

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

which is in the standard form:

$ax^2+bx+c = 0$

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

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = (color(blue)(0))^2-4(color(blue)(4))(color(blue)(9)) = 0-144 = -144$

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:

$A^2-B^2 = (A-B)(A+B)$

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

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

$0 = 4x^2+9$

$color(white)(0) = (2x)^2+3^2$

$color(white)(0) = (2x)^2-(3i)^2$

$color(white)(0) = (2x-3i)(2x+3i)$

Hence:

$x = +-3/2i$

How do you find the discriminant and how many and what type of solutions does 4x^2+ 9 = 04x^2+ 9 = 0 have?