Editing Quadratic formula (section)

== Derivation by completing the square ==
[[File:Quadratic formula via completing the square.png|thumb|upright=1.25|To complete the square, form a squared binomial on the left-hand side of a quadratic equation, from which the solution can be found by taking the square root of both sides.]]

The standard way to derive the quadratic formula is to apply the method of [[completing the square]] to the generic quadratic equation {{tmath|1=\textstyle ax^2 + bx + c = 0}}.<ref>{{citation | title=Schaum's Outline of Theory and Problems of Elementary Algebra | first1=Barnett | last1=Rich |first2=Philip |last2=Schmidt |publisher=The McGraw–Hill Companies | year=2004 |isbn=0-07-141083-X | url=https://books.google.com/books?id=8PRU9cTKprsC | at = [https://books.google.com/books?id=8PRU9cTKprsC&pg=PA291 Chapter 13 §4.4, p. 291]}}</ref><ref>Li, Xuhui. ''An Investigation of Secondary School Algebra Teachers' Mathematical Knowledge for Teaching Algebraic Equation Solving'', p. 56 (ProQuest, 2007): "The quadratic formula is the most general method for solving quadratic equations and is derived from another general method: completing the square."</ref><ref>Rockswold, Gary. ''College algebra and trigonometry and precalculus'', p. 178 (Addison Wesley, 2002).</ref><ref>Beckenbach, Edwin et al. ''Modern college algebra and trigonometry'', p. 81 (Wadsworth Pub. Co., 1986).</ref> The idea is to manipulate the equation into the form {{tmath|1=\textstyle (x + k)^2 = s}} for some expressions {{tmath|k}} and {{tmath|s}} written in terms of the coefficients; take the [[square root]] of both sides; and then isolate {{tmath|x}}.

We start by dividing the equation by the quadratic coefficient {{tmath|a}}, which is allowed because {{tmath|a}} is non-zero. Afterwards, we subtract the constant term {{tmath|c/a}} to isolate it on the right-hand side:

<math display=block>\begin{align}
ax^{2\vphantom|} + bx + c &= 0 \\[3mu]
x^2 + \frac{b}{a} x + \frac{c}{a} &= 0 \\[3mu]
x^2 + \frac{b}{a} x &= -\frac{c}{a}.
\end{align}</math>

The left-hand side is now of the form {{tmath|\textstyle x^2 + 2kx}}, and we can "complete the square" by adding a constant {{tmath|\textstyle k^2}} to obtain a squared binomial {{tmath|1=\textstyle x^2 + 2kx + k^2 = {} }}{{wbr}}{{tmath|\textstyle (x + k)^2}}. In this example we add {{tmath|\textstyle (b / 2a)^2}} to both sides so that the left-hand side can be factored (see the figure):

<math display=block>\begin{align}
x^2 + 2\left(\frac{b}{2a}\right)x + \left(\frac{b}{2a}\right)^2
&= -\frac{c}{a}+\left( \frac{b}{2a} \right)^2 \\[5mu]
\left(x + \frac{b}{2a}\right)^2
&= \frac{b^2 - 4ac}{4a^2} .
\end{align}</math>

Because the left-hand side is now a perfect square, we can easily take the square root of both sides:

<math display=block>
x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}.
</math>

Finally, subtracting {{tmath|b/2a}} from both sides to isolate {{tmath|x}} produces the quadratic formula:

<math display=block>
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} .
</math>