How do you solve the equation #6x^2-17x+12=0# by completing the square?

1 Answer
Sep 20, 2017

Answer:

#x = 3/2" "# and #" "x = 4/3#

Explanation:

To cut down on arithmetic involving fractions, multiply by #24 = 6*2^2# in order to make the leading term a perfect square and provide a factor #2# for the middle coefficient.

Once we reach a difference of squares, use the difference of squares identity:

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

with #A=(12x-17)# and #B=1#.

So:

#0 = 24(6x^2-17x+12)#

#color(white)(0) = 144x^2-408x+288#

#color(white)(0) = (12x)^2-2(12x)(17)+17^2-1#

#color(white)(0) = (12x-17)^2-1^2#

#color(white)(0) = ((12x-17)-1)((12x-17)+1)#

#color(white)(0) = (12x-18)(12x-16)#

#color(white)(0) = 6(2x-3)(4)(3x-4)#

#color(white)(0) = 24(2x-3)(3x-4)#

Hence roots:

#x = 3/2" "# and #" "x = 4/3#