Question 245854
{{{ax^2 + bx = -c}}}
Completing the square means we want one side of the equation to match left side of the pattern: {{{p^2 + 2pq + q^2 = (p+q)^2}}}. Matching this pattern is easier if the leading coefficient is 1 so I'll divide both sides by a:
{{{x^2 + (b/a)x = (-c)/a}}}
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 {{{(b/2a)^2}}}. This is the third term we need. So we will create it by adding it to both sides of the equation:
{{{x^2 + (b/a)x + (b/2a)^2 =  (b/2a)^2 + (-c)/a}}}
We now have the left side matching the pattern. So we can rewrite it as a perfect square:
{{{(x + b/2a)^2 =  (b/2a)^2 + (-c)/a}}}
(If you don't see this, multiply out the left side and see if you get {{{x^2 + (b/a)x + (b/2a)^2}}}). We have now completed the square. The rest is solving for x.<br>
On the right side we need to do some simplifying:
{{{(x + b/2a)^2 =  b^2/4a^2 + (-c)/a}}}
Get the denominators equal so we can add:
{{{(x + b/2a)^2 =  b^2/4a^2 + ((-c)/a)(4a/4a)}}}
{{{(x + b/2a)^2 =  b^2/4a^2 + (-4ac)/4a^2}}}
Add:
{{{(x + b/2a)^2 =  (b^2 + (-4ac))/4a^2}}}
or
{{{(x + b/2a)^2 =  (b^2 -4ac)/4a^2}}}
Now we can find the square root of each side:
{{{sqrt((x + b/2a)^2) =  sqrt((b^2 -4ac)/4a^2)}}}
On the right side we can simplify:
{{{sqrt((x + b/2a)^2) =  sqrt(b^2 -4ac)/sqrt(4a^2)}}}
{{{sqrt((x + b/2a)^2) =  sqrt(b^2 -4ac)/2a}}}
(We use {{{sqrt(4a^2) = 2a}}} instead of {{{sqrt(4a^2) = abs(2a)}}} 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(x + b/2a) =  sqrt(b^2 -4ac)/2a}}}
Solving absolute value equations requires two equations:
{{{x + b/2a =  sqrt(b^2 -4ac)/2a}}} or {{{x + b/2a =  -sqrt(b^2 -4ac)/2a}}}
Now we add -b/2a to each each side:
{{{x = -b/2a + sqrt(b^2 -4ac)/2a}}} or {{{x = - b/2a - sqrt(b^2 -4ac)/2a}}}
The two fractions on the right side of both equations conventiently have the same denominator so we can add them:
{{{x = (-b + sqrt(b^2 -4ac))/2a}}} or {{{x = (- b - sqrt(b^2 -4ac))/2a}}}
The shorthand for these two equations is:
{{{x = (-b +- sqrt(b^2 -4ac))/2a}}}
which is the familiar quadratic formula!