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#