Can someone show me how this is correct (completing the square...)?

Question

If y=Ax^2+Bx+C (,and A does not equal zero),

Then by completing the square you get:

(x+ (B / (2A)) )^2 = (1/A) ( y - C + ((B^2) / (4A)) )

My problem is that I don't understand how they were able to complete the square and get that answer.

Answer 1

I think you want to prove the Quadratic formula which is found at y=0

Ax^2+Bx+C=0

Put x's on one side

Ax^2+Bx= -C

Divide by A to get x^2 alone

x^2+(B/A)x=-C/A

When completing the square take (1/2) of the middle term=(B/A),  then square it=

1/4[B^2/A^2]

Then add that to both sides.:

x^2+(B/A)x+1/4[B^2/A^2] =-C/A+1/4[B^2/A^2]

Factoring gives [x+1/2(B/A)]^2=-C/A+1/4[B^2/A^2]

or rewriting as

[x+(B/(2A))]^2=[B^2/(4A^2)]-C/A

Get a common denominator for right side =4A^2, giving

[x+(B/(2A))]^2=[(B^2-4AC)/(4A^2)]

Take the sqrt of each side and solve for x to get:

x+B/(2A)= +-sqrt[(B^2-4AC)/(4A^2)]

Remember sqrt 4A^2=2A

Now we have x+B/(2A)= +-sqrt[(B^2-4AC)]/(2A)

Solving for x gives    

x= =+-sqrt[(B^2-4AC)]/(2A)-B/(2A)

Combining terms:

x={-B+-sqrt[(B^2-4AC)]}/(2A) the quadratic formula

Answer 2

AX^2 + BX + C = 0

X^2 + BX/2 = - C/A

X^2 + BX/2 + (B/2A)^2 = -C/A + (B/2A)^2

(X + B/2A)^2 = -C/A + (B/2A)^2 = -C/A + B^2/4A^2

(X + B/2A)^2 = (B^2 - 4AC) / 4A^2

X + B/2A = +- sqrt(B^2 - 4AC/4A^2)

X + B/2A = +- (sqrt(B^2 - 4AC)) / 2A

X = (-B +- sqrt(B^2 - 4AC) / 2A  

Answer 3

y=Ax^2+Bx+C

= A [ x^2 + (B/A)x + (C/A) ]

= A [ x^2 + 2(B/(2A))x + (C/A) ]

= A [ x^2 + 2((B/(2A))x + B^2/(2A)^2 + (C/A) - B^2/(2A)^2 ]

= A [ ( x + B/(2A) )^2 + (4C - B^2)/(4A^2) ]

(y/A) - A(4C - B^2)/(4A^2) = ( x + B/(2A) )^2

switching

( x + B/(2A) )^2 = (y/A) - (4C - B^2)/(4A)

or

( x + B/(2A) )^2 =[ 4y - 4C + B^2 ]/(4A)

a little bit different from what you have