Question

Complete the square to solve the equation for x: ax^2+bx+c=0. Provide a detailed account of your procedures.

Answer 1

`x = (-b + \sqrt{b^(2) - 4\, a\, c}) / (2\, a)` or
`x = (-b - \sqrt{b^(2) - 4\, a\, c}) / (2\, a)` assuming that
`b^(2) - 4\, a\, c \ge 0`.

Explanation:

Let
`h` and
`k` be constants. Consider
`a\, (x + h)^(2) = k`. In this equation,
`(x + h)^(2)`, the only term that includes
`x`, is a perfect square. If
`k \ge 0`, solving this equation is as simple as taking the square root of both sides of the equation:

`x + h = √(k / a)` or
`x + h = -√(k / a)`.

`x = (-h) + √(k / a)` or
`x = (-h) - √(k / a)`.

Assume that there are values for
`h` and
`k` such that
`a\, x^(2) + b\, x + c = 0` is equivalent to
`a\, (x + h)^(2) = k`. If
`(k / a) \ge 0`, then
`x = (-h) + √(k / a)` and
`x = (-h) - √(k / a)` would be solutions to
`a\, x^(2) + b\, x + c = 0\!`.

Apply binomial expansion to
`a\, (x + h)^(2) = k` and rewrite to find the values for
`h` and
`k`:

`a\, (x^(2) + 2\, h\, x + h^(2)) - k = 0`.

`a\, x^(2) + 2\, a\, h\, x + (a\, h^(2) - k) = 0`.

Match the coefficients of this equation with those in
`a\, x^(2) + b\, x + c = 0`:

`2\, a\, h = b`.

`a\, h^(2) - k = c`.

Solve for
`h` and
`k` in terms of
`a`,
`b`, and
`c`:

`h = (b / 2\, a)`.

`\begin{aligned}k &= a\, h^(2) - c \\ &= (a\, b^(2))/(4\, a^(2)) - c \\ &= (b^(2))/(4\, a) - c \\ &= (b^(2) - 4\, a\, c)/(4\, a)\end{aligned}`.

Hence, as long as
`(b^(2) - 4\, a\, c) \ge 0`, (such that
`(k / a) \ge 0`,) solutions to
`a\, x^(2) + b\, x + c = 0` would be:

`\begin{aligned}x &= (-h) + \sqrt{(k)/(a)} \\ &= -(b)/(2\, a) + \sqrt{(b^(2) - 4\, a\, c)/(4\, a^(2))} \\ &= -(b)/(2\, a) + \frac{\sqrt{b^(2) - 4\, a\, c}}{2\, a}\end{aligned}`, and

`\begin{aligned}x &= (-h) - \sqrt{(k)/(a)} \\ &= -(b)/(2\, a) - \sqrt{(b^(2) - 4\, a\, c)/(4\, a^(2))} \\ &= -(b)/(2\, a) - \frac{\sqrt{b^(2) - 4\, a\, c}}{2\, a}\end{aligned}`.