How do you solve #x^2 + 24x + 90 = 0#?

1 Answer
Mar 28, 2016

Answer:

Complete the square to find:

#x = -12+-3sqrt(6)#

Explanation:

This can be solved by completing the square.

Also use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x+12)# and #b=3sqrt(6)# as follows:

#0 = x^2+24x+90#

#= (x+12)^2-144+90#

#= (x+12)^2-54#

#= (x+12)^2-(3^2*6)#

#= (x+12)^2-(3sqrt(6))^2#

#= ((x+12)-3sqrt(6))((x+12)+3sqrt(6))#

#= (x+12-3sqrt(6))(x+12+3sqrt(6))#

Hence:

#x = -12+-3sqrt(6)#