I need help completing the square of equations in Algebra II.?

3x^2 + x - 2 = 0 <--------> a(x^)2 + b(x) + constant = 0

I actually like to set the equation equal to the constant (only when completing the square).

3(x^2) + x = 2

Set the x^2 coefficient equal to 1. This can be down by dividing the entire equation by 1 / 3.

[3(x^2) + x](1 / 3) = 2 (1 / 3)

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

Take the b term and multiply by 1 / 2, and then square it:

[(b)(1 / 2)]^2 = [(1 / 3)(1 / 2)]^2 = ( 1/ 6)^2 = (1 / 6)(1 / 6) = 1 / 36

Now add this value to both sides of the equation:

x^2 + (1 / 3)(x) + 1 / 36 = (2 / 3) + (1 / 36)

The left side of the equation automatically factors to this:

(x + b / 2)^2 = __

(x + 1 / 6)^2 = 2 / 3 + 1 / 36

Now I will need a common denominator of 36:

(x + 1 / 6)^2 = (2 / 3)(12 / 12) + ( 1 / 36)

(x + 1 / 6)^2 = 24 / 36 + 1 / 36

(x + 1 / 6)^2 = 25 / 36

Now take the square root of both sides of the equation to cancel the squaring:

√[(x + 1/ 6)^2 = √(25 / 36)

x + 1 / 6 = √25 / √36

Subtract the 1 / 6 to both sides of the equation:

x = ± 5 / 6 - 1 / 6

x = (-1 ± 5) / 6

You get one value when you add the 5:

x = 4 / 6 -----> reduce by halve: x = 4(1/2) / 6(1/2) ---------> x = 2 / 3

You get another value when you subtract the 5:

x = -1

---------------

2x^2 + 8x - 3 = 0

Remember the conformation i like:

2x^2 + 8x = 3

Divide all terms by 2:

x^2 + 4x = 3 / 2

Add a new term in: (b / 2)^2 = (4 / 2)^2 = (2)^2 = 4

x^2 + 4x + 4 = 3 / 2 + 4

this factors rather intuitively:

(x + 2)^2 = (3 / 2) + (4 / 1)( 2 / 2 )

(x + 2)^2 = (3 / 2) + 8 / 2

(x + 2)^2 = (3 + 8) / 2

Take the square root of both sides:

√[(x + 2)^2 = ± √(11 / 2)

Distribute the square root:

x + 2 = (± √11 / √2)

isolate x:

x = ± (√11 / √2) - 2

We cannot tolerate a square root in the denominator:

(√11 / √2)( √2 / √2) = (√11)(√2) / (√2 * √2) = √(11 * 2) / √4 -------> √22 / 2

Perform that substitution:

x = ± (√22 / 2) - 2

x = ± (√22 / 2) - (2 / 1)(2 / 2)

x = ± (√22 / 2) - 4 / 2

x = (- 4 ± √22) / 2

Here are the approximations:

x = 0.3452

x = -4.3452

{sqrt(3)*x + 1/[2*sqrt(3)]}^2 = (25/12)

[sqrt(2)*x + 4/sqrt(2)]^2 = 5

Basically you want to make sure when you FOIL it out, the x terms equal your original equation's x terms, and you can add or subtract the constant to make the whole equation equal to the original.

{sqrt(a)*x + b/[2*sqrt(a)]}^2 - b/[2*sqrt(a)] + c = 0

3x^2 + x -- 2 = 0

OR x^2 + x/3 -- 2/3 = 0 , on dividing given equation 3x^2 + x -- 2 = 0 by 3.

OR x^2 + 2x(1/6) + (1/6)^2 = 2/3 + (1/6)^2

OR (x + 1/6)^2 = (5/6)^2

OR x + 1/6 = 5/6, --5/6

OR x = 4/6, --6/6 = 2/3, --1 ANSWER

2x^2 + 8x -- 3 = 0

OR x^2 + 4x -- 3/2 = 0 , on dividing given equation 2x^2 + 8x -- 3 = 0 by 2.

OR x^2 + 2x(2) + (2)^2 = 3/2 + (2)^2

OR (x + 2)^2 = (11/2)

OR x + 2 = +/-- sqrt(11/2)

OR x = -- 2 +/-- sqrt(11/2) ANSWER