How do you solve #8y^2 - 22y + 15 = 0#?

1 Answer
Jun 12, 2015

Answer:

Use a modified AC Method to find: #8y^2-22y+15 = (2y-3)(4y-5)#, hence #y=3/2# or #y=5/4#.

Explanation:

#f(y) = 8y^2-22y+15# is of the form #ay^2+by+c# with #a=8#, #b=-22# and #c=15#.

Let #A=8#, #B=22# and #C=15#.

Look for a factorization of #AC=8*15=120# into a pair of factors whose sum is #B=22#.

Notice that #A+C=23#, which is close to #22#, so the pair of factors we are looking for is close to #8, 15#. Notice that #B1=12# and #B2=10# works.

Then for each pair #(A, B1)# and #(A, B2)#, divide by the HCF (highest common factor) to get a pair of coefficients of a factor of #f(y)# (choosing suitable signs) as follows:

#(A, B1) = (8, 12) -> (2, 3) -> (2y-3)#
#(A, B2) = (8, 10) -> (4, 5) -> (4y-5)#

Hence #8y^2-22y+15 = (2y-3)(4y-5)#

If #f(y) = 0# then #(2y-3) = 0# giving #y = 3/2#

or #(4y-5) = 0# giving #y = 5/4#.