How do you solve #2x ^ { 2} - 12x + 18= 0#?

1 Answer
Aviv S.
Apr 30, 2018

The only solution is #x=3#.

Explanation:

First, divide everything by #2#.

#2x^2-12x+18=0#

#x^2-6x+9=0#

Then, find two numbers that multiply to #9# (the last number) and add up to #-6# (the middle number). These two numbers are #-3# and #-3#. Split the #x# terms up into these values:

#x^2-3x-3x+9=0#

Now factor the first two and last two terms separately:

#color(red)x(x-3)-3x+9=0#

#color(red)x(x-3)-color(blue)3(x-3)=0#

#(color(red)x-color(blue)3)(x-3)=0#

#(x-3)^2=0#

#x-3=sqrt0#

#x-3=0#

#x=3#

That's the only solution. Hope this helped!

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