How do you solve the quadratic #6y^2-13y+6=0# using any method?

2 Answers
Oct 6, 2016

Answer:

Using the quadratic formula
#color(white)("XXX")color(green)(y=2/3color(black)(" or ")y=3/2)#

Explanation:

The quadratic formula for the general quadratic (in #y#)
#color(white)("XXX")ay^2+by+c=0#
is
#color(white)("XXX")y=(-b+-sqrt(b^2+4ac))/(2a)#

For the example: #6y^2-13y+6=0#
#color(white)("XXX")a=6#
#color(white)("XXX")b=-13#
#color(white)("XXX")c=6#
giving
#color(white)("XXX")y=(13+-sqrt((-13)^2-4(6)(6)))/(2(6))#

#color(white)("XXX")=(13+-sqrt(169-144))/12#

#color(white)("XXX")=(13+-5)/12#

#color(white)("XXX")=18/12= 3/2color(black)(" or ")=8/12=2/3#

Oct 6, 2016

Answer:

#y = 2/3# or #y = 3/2#

Explanation:

Use an AC method:

Look for a pair of factors of #AC = 6*6=36# with sum #B=13#

The pair #9, 4# works.

Use this pair to split the middle term, then factor by grouping as follows:

#0 = 6y^2-13y+6#

#color(white)(0) = 6y^2-9y-4y+6#

#color(white)(0) = (6y^2-9y)-(4y-6)#

#color(white)(0) = 3y(2y-3)-2(2y-3)#

#color(white)(0) = (3y-2)(2y-3)#

Hence #y = 2/3# or #y = 3/2#