Question

Solve by completing the square. 3x^2-6x=2 Solve by completing the square. 3x^2-6x=2 @Mathematics

Answer 1

\[x^2-2x-\frac{2}{3}=0\]\[(x-\frac{2}{2})^2 - (\frac{2}{2})^2 -\frac{2}{3} =0\]\[(x-1)^2-1-\frac{2}{3} =0\]\[(x-1)^2=\frac{5}{3}\]\[x-1 = \pm \sqrt{\frac{5}{3}}\]\[x=1 \pm \sqrt{\frac{5}{3}}\]

Answer 2

Equation becomes : 3x^2-6x-2=0
here , a = 3
b = -6
c = -2
divide whole equation by a (3)
Equation becomes = x^2 - 2x - 2/3 = 0
=> x^2 - 2x = 2/3
Add (b/2a)^2 to both sides..
=> x^2-2x +(3/2)^2 = 2/3 + (3/2)^2
=> (x- 3/2)^2 = 2/3 + 9/4
=> x = SQRT(2/3+9/4) + 3/2
Calculate it :)

Answer 3

\[x^2 - bx +c =0\]completin the square
\[(x-\frac{b}{2})^2 - (\frac{b}{2})^2 +c=0\]
This is what i did ^

Answer 4

Thank you for your help you are always awesome.

Answer 5

ur welcome:D

Answer 6

why can't you just use the quadratic formula, which is derived from completing the square?