Question 231170
Your equation is:


3x^2 - 6x = 3


You need to divide both sides of this equation by 3 to get:


x^2 - 2x = 1


You take 1/2 of 2 to get 1.


Your factors are (x-1)^2 = 1 + 1


This becomes (x-1)^2 = 2


Take the square root of this to get:


x-1 = +/- sqrt(2)


Add 1 to both sides to get:


x = 1 +/- sqrt(2)


This means that:


x = 1+sqrt(2)
or:
x = 1-sqrt(2)


To see if this works out ok, you need to substitute in the original equation to see if the equation is true.


Your original equation is 3x^2 - 6x = 3


If x = 1 + sqrt(2), this becomes:


3 * (1+sqrt(2)^2) - 6 * (1+sqrt(2)) = 3


This becomes:


3 + 6*sqrt(2) + 6 - 6 - 6*sqrt(2) = 3


6 and 6*sqrt(2) cancel out leaving 3 = 3


This is true so the value for x is good.


I'll leave you to replace x with 1 - sqrt(2) to see if that's good too.


You might want to check out the following website to review how to complete the square.


<a href = "http://www.algebrahelp.com/lessons/equations/completingthesquare/" target = "_blank">Completing the Square (algebrahelp)</a>