How do you solve #18u^2-3u=1#?

2 Answers
Jun 28, 2015

Answer:

#u=1/3# or #u= -1/6#

Explanation:

Method 1
Solutions for #18u^2-3u = 1# or equivalently #18u^2-3u -1 =0#
can be determined using the quadratic formula (see bottom, if you don't know this formula).

#u= (-(-3)+-sqrt((-3)^2-4(18)(-1)))/(2(18))#

#color(white)("XXXX")##= (3+-sqrt(81))/(36)#

#u = 12/36 = 1/3#
or
#u=-6/36 = -1/6#

Method 2
If we recognize that (after moving the 1 to the left side) we can factor:
#color(white)("XXXX")##18y^2-3u-1=0#
to get
#color(white)("XXXX")##(3u-1)(6u+1)=0#

Then either
#color(white)("XXXX")##(3u-1) = 0# which implies #u=1/3#
or
#color(white)("XXXX")##(6u+1)=0# which implies #u=-1/6#

Jun 28, 2015

Answer:

Factor #y = 18x^2 - 3x - 1#

Explanation:

#y = 18x^2 - 3x - 1 = 18(x - p)(x - q).#
I use the new AC Method (Google, Yahoo Search)
Converted #y' = x^2 - 3x - 18# = (x - p')(x - q').
Factor pairs of (-18)--> (-2, 9)(-3, 6). This sum is 3 = -b
p' = 3 and q' = - 6
#p = 3/18 = 1/6# and #q = -6/18 = - 1/3#

Factored form: y = 18(x + 1/6)(x - 1/3) =# (6x + 1)(3x - 1)#