How do you solve 0=3x^2-2x-12 by completing the square?

2 Answers
May 16, 2015

In general ax^2+bx+c = a(x+b/(2a))^2+(c-b^2/(4a))

For your problem, a=3, b=-2 and c=-12, so we get

0 = ax^2+bx+c = 3x^2 - 2x - 12

= 3(x+(-2)/(2*3))^2+(-12-(-2)^2/(4*3))

= 3(x-2/6)^2-(12+4/12)

= 3(x-1/3)^2-(12+1/3)

= 3(x-1/3)^2-37/3

Adding 37/3 to both sides we get

3(x-1/3)^2 = 37/3

Dividing both sides by 3 we get

(x-1/3)^2 = 37/(3^2)

Hence

x-1/3 = +- sqrt(37/3^2) = +- sqrt(37)/3

Adding 1/3 to both sides, we get

x = 1/3 +- sqrt(37)/3 = (1 +- sqrt(37))/3

May 16, 2015

If 3x^2-2x-12 = 0
then
x^2-2/3x -4 = 0

isolate the constant
x^2-2/3x = 4

complete the square
x^2 - 2/3x + (1/3)^2 = 4 + (1/3)^2

(x-1/3)^2 = 37/9

take the square root
x-1/3 = +-sqrt(37)/3

simplify
x= (1+-sqrt(37))/3