Question

3x^2+3x+2y=0 complete the square and reduce to one standard form y-b=A(x-a)^2 or x-a=A(y-b)^2

Answer 1

\[3x^2+3x+2y=0\]
\[3(x^2+x)=-2y\]
\[3[x^2+x + (\frac{1}{2})^2-(\frac{1}{2})^2]=-2y\]\[3[x^2+x + (\frac{1}{2})^2-(\frac{1}{2})^2]=-2y\]\[3(x+\frac{1}{2})^2-\frac{3}{4}=-2y\]Divide both sides by -2\[-\frac{3}{2}(x+\frac{1}{2})^2+\frac{3}{8}=y\]\[-\frac{3}{2}(x+\frac{1}{2})^2=y-\frac{3}{8}\]\[y-\frac{3}{8}=-\frac{3}{2}(x+\frac{1}{2})^2\]

Answer 2

why did u squared 1/2

Answer 3

i think u should add 1/2 on both sides

Answer 4

Because of the identity
\((a+b)^2 = a^2+2ab+b^2\)
For the expression x^2 + x , a=x, 2ab = x
So, 2xb = x
b = 1/2

To complete square, we need to add b^2, that is (1/2)^2 = 1/4
But we can't just add 1/4. So, I add it and subtract it, like this:
\[3[x^2+x + (\frac{1}{2})^2-(\frac{1}{2})^2]=-2y\]

Actually, adding a term to both sides is EQUAL to adding the term and subtracting the same term on the SAME side.
Like this:
a = 5 , assume we add b to both sides
a + b = 5 + b
a + b - b = 5
See ?

Answer 5

and last question how did u get (1/2)^2 = 3/4

Answer 6

IN 4TH STEP

Answer 7

There are 2 step to get that value.
\[3[x^2+x + (\frac{1}{2})^2-(\frac{1}{2})^2]=-2y\]\[3(x^2+x + (\frac{1}{2})^2)-3(\frac{1}{2})^2=-2y\]\[3(x+\frac{1}{2})^2-3(\frac{1}{4})=-2y\]\[3(x+\frac{1}{2})^2-(\frac{3}{4})=-2y\]

Answer 8

where did u get that -3

Answer 9

Distributive.

Answer 10

hoooo u got got it from the first 3