SOLUTION: Prove the quadratic formula by completing the square. ax^2+by=-c

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Question 245854: Prove the quadratic formula by completing the square.
ax^2+by=-c

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
ax%5E2+%2B+bx+=+-c
Completing the square means we want one side of the equation to match left side of the pattern: p%5E2+%2B+2pq+%2B+q%5E2+=+%28p%2Bq%29%5E2. Matching this pattern is easier if the leading coefficient is 1 so I'll divide both sides by a:
x%5E2+%2B+%28b%2Fa%29x+=+%28-c%29%2Fa
Now comes the tricky part. We need to figure what third term we need on the left side to match the pattern. With a leading coefficient of 1, this is a little easier. The third term we need is the square of 1/2 of the coefficient of x. The coefficient of x is b/a and 1/2 of this is b/2a and the square of b/2a is %28b%2F2a%29%5E2. This is the third term we need. So we will create it by adding it to both sides of the equation:
x%5E2+%2B+%28b%2Fa%29x+%2B+%28b%2F2a%29%5E2+=++%28b%2F2a%29%5E2+%2B+%28-c%29%2Fa
We now have the left side matching the pattern. So we can rewrite it as a perfect square:
%28x+%2B+b%2F2a%29%5E2+=++%28b%2F2a%29%5E2+%2B+%28-c%29%2Fa
(If you don't see this, multiply out the left side and see if you get x%5E2+%2B+%28b%2Fa%29x+%2B+%28b%2F2a%29%5E2). We have now completed the square. The rest is solving for x.

On the right side we need to do some simplifying:
%28x+%2B+b%2F2a%29%5E2+=++b%5E2%2F4a%5E2+%2B+%28-c%29%2Fa
Get the denominators equal so we can add:
%28x+%2B+b%2F2a%29%5E2+=++b%5E2%2F4a%5E2+%2B+%28%28-c%29%2Fa%29%284a%2F4a%29
%28x+%2B+b%2F2a%29%5E2+=++b%5E2%2F4a%5E2+%2B+%28-4ac%29%2F4a%5E2
Add:
%28x+%2B+b%2F2a%29%5E2+=++%28b%5E2+%2B+%28-4ac%29%29%2F4a%5E2
or
%28x+%2B+b%2F2a%29%5E2+=++%28b%5E2+-4ac%29%2F4a%5E2
Now we can find the square root of each side:
sqrt%28%28x+%2B+b%2F2a%29%5E2%29+=++sqrt%28%28b%5E2+-4ac%29%2F4a%5E2%29
On the right side we can simplify:
sqrt%28%28x+%2B+b%2F2a%29%5E2%29+=++sqrt%28b%5E2+-4ac%29%2Fsqrt%284a%5E2%29
sqrt%28%28x+%2B+b%2F2a%29%5E2%29+=++sqrt%28b%5E2+-4ac%29%2F2a
(We use sqrt%284a%5E2%29+=+2a instead of sqrt%284a%5E2%29+=+abs%282a%29 because we will end up with both the positive and negative values of the right side anyway.) Now we can simplify the left side:
abs%28x+%2B+b%2F2a%29+=++sqrt%28b%5E2+-4ac%29%2F2a
Solving absolute value equations requires two equations:
x+%2B+b%2F2a+=++sqrt%28b%5E2+-4ac%29%2F2a or x+%2B+b%2F2a+=++-sqrt%28b%5E2+-4ac%29%2F2a
Now we add -b/2a to each each side:
x+=+-b%2F2a+%2B+sqrt%28b%5E2+-4ac%29%2F2a or x+=+-+b%2F2a+-+sqrt%28b%5E2+-4ac%29%2F2a
The two fractions on the right side of both equations conventiently have the same denominator so we can add them:
x+=+%28-b+%2B+sqrt%28b%5E2+-4ac%29%29%2F2a or x+=+%28-+b+-+sqrt%28b%5E2+-4ac%29%29%2F2a
The shorthand for these two equations is:
x+=+%28-b+%2B-+sqrt%28b%5E2+-4ac%29%29%2F2a
which is the familiar quadratic formula!