How do you solve #3x ^ { 2} - 1= - 2x#?

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Answer 1 · 2017-03-15T13:56:38

See the entire solution process below:

FIrst, add #color(red)(2x)# to each side of the equation to put this equation in standard form:

#3x^2 - 1 + color(red)(2x) = -2x + color(red)(2x)#

#3x^2 + color(red)(2x) - 1 = 0#

#3x^2 + 2x - 1 = 0#

Next, factor this as:

#(3x - 1)(x + 1) = 0

Now, solve each term for #0#:

Solution 1)

#3x - 1 = 0#

#3x - 1 + color(red)(1) = 0 + color(red)(1)#

#3x - 0 = 1#

#3x = 1#

#(3x)/color(red)(3) = 1/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 1/3#

#x = 1/3#

Solution 2)

#x + 1 = 0#

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

#x + 0 = -1#

#x = -1#

The solution is: #x = 1/3# and #x = -1#