Completing the square

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Very simply, given a problem of the form:

We may transform the equation such that we no loger have both "x" and "x squared" with a method called "completing the square". To do this we add one half of the coefficient of "x squared" to both sides of the equation, like so:

<math>\begin{matrix}

x^2 + bx &=& c \\ x^2 + bx + (\frac{b}{2})^2 &=& c+ (\frac{b}{2})^2 \\ (x + \frac{b}{2})^2 &=& c + (\frac{b}{2})^2 \\ (x + \frac{b}{2}) &=& \sqrt{c + (\frac{b}{2})^2} \end{matrix}</math>

More specifically, Completing the square is a technique of elementary algebra wherein an expression

is replaced by one of the form

Specifically, we have:

<math>ax^2 + bx + c = a\left(x^2 + \frac{bx}{a}\right) +c </math>

<math>= a\left(x^2 + \frac{bx}{a} + \left(\frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right)\right) + c </math>

<math>= a\left(x^2+2\frac{bx}{2a}+\left(\frac{b}{2a}\right)^2\right)-\frac{b^2}{4a} +c </math>

<math>= a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a} + c.</math>

Completing the square reduces any problem involving a quadratic polynomial to one involving a square quadratic polynomial and a constant.

See also: quadratic equation

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Example

A very simple example is:

<math>\begin{matrix}

x^2 + 6x &=& 16 \\ x^2 + 6x + (\frac{6}{2})^2 &=& 16 + (\frac{6}{2})^2 \\ x^2 + 6x + 9 &=& 16 + 9 \\ (x + 3)^2 &=& 25 \\ (x + 3) &=& \sqrt{25} \\ x + 3 &=& \pm5 \\ x &=& \pm5 - 3 \\ \end{matrix}</math>

Another fairly simple example is:

Now, consider the problem of finding this antiderivative:

The denominator is

Adding (10/2)² = 25 to x² - 10x gives a perfect square x² - 10x + 25 = (x - 5)². So we get

Let our integral be

<math>\int\frac{dx}{9x^2-90x+241}=\frac{1}{9}\int\frac{dx}{(x-5)^2+(4/3)^2}=\frac{1}{9}\cdot\frac{3}{4}\arctan\frac{3(x-5)}{4}+C.</math>

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