If I use the method of completing the square on
3w^2-4w+1=0
How would I solve??
w^2-4/3 w = -1
ok, take half the b term, and square it.
((-4/3)*1/2)^2= (-2/3)^2= 4/9
add that to each side.
w^2-4/3 w + 4/9=-1+4/9
(w-2/3)^2=-5/9
(w-2/3)^2-(5/9)=0
which now you have a difference of two squares..
(w-2/3)^2-(1/3 sqrt5)^2=0
and you factor that as the difference of two squares
(w-2/3+1/3 sqrt5)(w-2/3 -1/3 sqrt 5)=0
bob forgot to divide the right side by 3
w^2-4/3 w = -1/3
w^2-4/3 w + 4/9= -1 + 4/9
(x-2/3)^2 = 1/9
x - 2/3 = ±1/3
x = 1 or x = 1/3
I agree. Duh.
To solve the quadratic equation 3w^2 - 4w + 1 = 0 using the method of completing the square, follow these steps:
Step 1: Move the constant term to the right side of the equation.
3w^2 - 4w = -1
Step 2: Divide through by the coefficient of the squared term.
w^2 - (4/3)w = -1/3
Step 3: Take half the coefficient of the linear term and square it. Add this new term to both sides of the equation.
w^2 - (4/3)w + (2/3)^2 = -1/3 + (2/3)^2
Simplifying the right side:
w^2 - (4/3)w + (2/3)^2 = -1/3 + 4/9
w^2 - (4/3)w + (2/3)^2 = -3/9 + 4/9
w^2 - (4/3)w + (2/3)^2 = 1/9
Step 4: Rewrite the left side as a perfect square.
(w - 2/3)^2 = 1/9
Step 5: Take the square root of both sides.
√[(w - 2/3)^2] = ±√(1/9)
Simplifying the right side:
w - 2/3 = ±1/3
Step 6: Solve for w.
When w - 2/3 = 1/3, w = 1/3 + 2/3 = 1
When w - 2/3 = -1/3, w = -1/3 + 2/3 = 1/3
So the solutions to the equation 3w^2 - 4w + 1 = 0 are w = 1 and w = 1/3.