How do you factor and solve #6x^2 - 5x = 6#?

2 Answers
Aug 21, 2016

Answer:

#x=-2/3# or #x=3/2#

Explanation:

First subtract #6# from both sides to get:

#6x^2-5x-6 = 0#

Factor using an AC method:

Find a pair of factors of #AC = 6*6=36# which differ by #B=5#.

The pair #9, 4# works. Use this pair to split the middle term, then factor by grouping:

#0 = 6x^2-5x-6#

#=6x^2-9x+4x-6#

#=(6x^2-9x)+(4x-6)#

#=3x(2x-3)+2(2x-3)#

#=(3x+2)(2x-3)#

Hence zeros: #x=-2/3# and #x=3/2#

Aug 21, 2016

Answer:

The solutions are the values of #x# that satisfy #6x^2-5x-6=0 #

Explanation:

By completing the square you can get the formula for the solutions (roots) of a 2nd degree polynomial equation #ax^2+bx+c=0#, as follows #x_1,_2 = (-b+-sqrt(b^2-4ac))/(2a)#. In this case we have #a=6#, #b=-5# and #c=-6#

So the first solution is #x_1 = (-b+sqrt(b^2-4ac))/(2a)=(5+sqrt(25-4*6*(-6)))/(2*6)=(5+sqrt(25+144))/(12)=(5+sqrt(169))/(12)=(5+13)/12=18/12=3/2#

Similarly, the second solution is #(5-13)/12=-8/12=-2/3#