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

1 Answer

7 votes

Answer:


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}.

User Jetpackpony
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