What are the zero(s) of: #3x^2 - 5x -4 = 0#?

1 Answer
May 22, 2018

Answer:

#x = 5/6+-sqrt(73)/6#

Explanation:

Given:

#3x^2-5x-4 = 0#

The difference of squares identity can be written:

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

Complete the square and use this with #A=(6x-5)# and #B=sqrt(73)# as follows:

#0 = 12(3x^2-5x-4)#

#color(white)(0) = 36x^2-60x-48#

#color(white)(0) = (6x)^2-2(6x)(5)+25-73#

#color(white)(0) = (6x-5)^2-(sqrt(73))^2#

#color(white)(0) = ((6x-5)-sqrt(73))((6x-5)+sqrt(73))#

#color(white)(0) = (6x-5-sqrt(73))(6x-5+sqrt(73))#

Hence:

#6x = 5+-sqrt(73)#

Hence:

#x = 5/6+-sqrt(73)/6#