How do you solve #12x^2 - 6x = 0#?

1 Answer
Mar 13, 2018

Answer:

See a solution process below:

Explanation:

First, factor an #6x# from each term on the left side of the equation:

#(6x * 2x) - (6x * 1) = 0#

#6x(2x - 1) = 0#

Now, solve each term on the left for #0#:

Solution 1:

#6x = 0#

#(6x)/color(red)(6) = 0/color(red)(6)#

#(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = 0#

#x = 0#

Solution 2:

#2x - 1 = 0#

#2x - 1 + color(red)(1) = 0 + color(red)(1)#

#2x - 0 = 1#

#2x = 1#

#(2x)/color(red)(2) = 1/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 1/2#

#x = 1/2#

The Solution Set Is: #x = {0, 1/2}#