How do you solve using the completing the square method #-3x^2-5x+5 = 33 #?

1 Answer
Mar 4, 2017

Answer:

#x = -5/6+-sqrt(311)/6i#

Explanation:

Given:

#-3x^2-5x+5=33#

Adding #3x^2+5x-5# to both sides and transposing, we get:

#3x^2+5x+28 = 0#

This is in standard form:

#ax^2+bx+c = 0#

with #a=3#, #b=5# and #c=28#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 5^2-4(3)(28) = 25-336 = -311#

Since #Delta < 0#, this quadratic has no Real solutions.

We can find Complex solutions by completing the square, then using the difference of squares identity:

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

with #A=(6x+5)# and #B=sqrt(311)# as follows:

#0 = 12(3x^2+5x+28)#

#color(white)(0) = 36x^2+60x+336#

#color(white)(0) = (6x)^2+2(6x)(5)+5^2+311#

#color(white)(0) = (6x+5)^2+(sqrt(311))^2#

#color(white)(0) = (6x+5)^2-(sqrt(311)i)^2#

#color(white)(0) = ((6x+5)-sqrt(311)i)((6x+5)+sqrt(311)i)#

#color(white)(0) = (6x+5-sqrt(311)i)(6x+5+sqrt(311)i)#

Hence:

#x = 1/6(-5+-sqrt(311)i) = -5/6+-sqrt(311)/6i#