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| [[File:Completing the square.ogv|thumb|right|400px|Animation depicting the process of completing the square. ([[:File:Completing the square.ogv|Details]], [[:File:Completing the square.gif|animated GIF version]])]]
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| In [[elementary algebra]], '''completing the square''' is a technique for converting a [[quadratic polynomial]] of the form
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| :<math>ax^2 + bx + c\,\!</math>
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| to the form
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| : <math> a(\cdots\cdots)^2 + \mbox{constant}.\, </math>
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| In this context, "constant" means not depending on ''x''. The expression inside the parenthesis is of the form (''x'' + constant). Thus
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| :<math>ax^2 + bx + c\,\!</math> is converted to
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| : <math> a(x + h)^2 + k\, </math>
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| for some values of ''h'' and ''k''.
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| Completing the square is used in
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| * solving [[quadratic equation]]s,
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| * graphing [[quadratic function]]s,
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| * evaluating [[integral]]s in calculus, such as [[Gaussian Integral|Gaussian integrals]] with a linear term in the exponent
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| * finding [[Laplace transforms]].
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| In mathematics, completing the square is considered a basic algebraic operation, and is often applied without remark in any computation involving quadratic polynomials. Completing the square is also used to derive the [[quadratic formula]].
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| ==Overview==
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| ===Background===
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| There is a simple formula in [[elementary algebra]] for computing the [[square (algebra)|square]] of a [[binomial]]:
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| :<math>(x + p)^2 \,=\, x^2 + 2px + p^2.\,\!</math>
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| For example:
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| :<math>\begin{alignat}{2}
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| (x+3)^2 \,&=\, x^2 + 6x + 9 && (p=3)\\[3pt]
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| (x-5)^2 \,&=\, x^2 - 10x + 25\qquad && (p=-5).
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| \end{alignat}
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| </math>
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| In any perfect square, the number ''p'' is always half the [[coefficient]] of ''x'', and the [[constant term]] is equal to ''p''<sup>2</sup>.
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| ===Basic example===
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| Consider the following quadratic [[polynomial]]:
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| :<math>x^2 + 10x + 28.\,\!</math>
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| This quadratic is not a perfect square, since 28 is not the square of 5:
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| :<math>(x+5)^2 \,=\, x^2 + 10x + 25.\,\!</math>
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| However, it is possible to write the original quadratic as the sum of this square and a constant:
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| :<math>x^2 + 10x + 28 \,=\, (x+5)^2 + 3.</math>
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| This is called '''completing the square'''.
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| ===General description===
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| Given any [[Monic polynomial|monic]] quadratic
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| :<math>x^2 + bx + c,\,\!</math>
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| it is possible to form a square that has the same first two terms:
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| :<math>\left(x+\tfrac{1}{2} b\right)^2 \,=\, x^2 + bx + \tfrac{1}{4}b^2.</math>
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| This square differs from the original quadratic only in the value of the constant
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| term. Therefore, we can write
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| :<math>x^2 + bx + c \,=\, \left(x + \tfrac{1}{2}b\right)^2 + k,</math>
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| where ''k'' is a constant. This operation is known as '''completing the square'''.
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| For example:
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| :<math>\begin{alignat}{1}
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| x^2 + 6x + 11 \,&=\, (x+3)^2 + 2 \\[3pt]
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| x^2 + 14x + 30 \,&=\, (x+7)^2 - 19 \\[3pt]
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| x^2 - 2x + 7 \,&=\, (x-1)^2 + 6.
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| \end{alignat}
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| </math>
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| ===Non-monic case===
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| Given a quadratic polynomial of the form
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| :<math>ax^2 + bx + c\,\!</math>
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| it is possible to factor out the coefficient ''a'', and then complete the square for the resulting [[monic polynomial]].
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| Example:
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| :<math>
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| \begin{align}
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| 3x^2 + 12x + 27 &= 3(x^2+4x+9)\\
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| &{}= 3\left((x+2)^2 + 5\right)\\
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| &{}= 3(x+2)^2 + 15
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| \end{align}</math>
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| This allows us to write any quadratic polynomial in the form
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| :<math>a(x-h)^2 + k.\,\!</math>
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| ===Formula===
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| The result of completing the square may be written as a formula. For the general case:<ref>{{cite book
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| |title=Precalculus: Building Concepts and Connections
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| |first1=Revathi
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| |last1=Narasimhan
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| |publisher=Cengage Learning
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| |year=2008
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| |isbn=0-618-41301-4
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| |pages=133–134
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| |url=http://books.google.com/books?id=hLZz3xcP0SAC}}, [http://books.google.com/books?id=hLZz3xcP0SAC&pg=PA134 Section ''Formula for the Vertex of a Quadratic Function'', page 133–134, figure 2.4.8]
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| </ref>
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| :<math>ax^2 + bx + c \;=\; a(x-h)^2 + k,\quad\text{where}\quad h = -\frac{b}{2a} \quad\text{and}\quad k = c - \frac{b^2}{4a}.</math>
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| Specifically, when ''a''=1:
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| :<math>x^2 + bx + c \;=\; (x-h)^2 + k,\quad\text{where}\quad h = -\frac{b}{2} \quad\text{and}\quad k = c - \frac{b^2}{4}.</math>
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| The matrix case looks very similar:
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| :<math>x^{\mathrm{T}}Ax + x^{\mathrm{T}}b + c = (x - h)^{\mathrm{T}}A(x - h) + k \quad\text{where}\quad h = -\frac{1}{2}A^{-1}b \quad\text{and}\quad k = c - \frac{1}{4}b^{\mathrm{T}}A^{-1}b</math>
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| where <math>A</math> has to be symmetric.
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| If <math>A</math> is not symmetric the formulae for <math>h</math> and <math>k</math> have
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| to be generalized to:
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| :<math>h = -(A+A^{\mathrm{T}})^{-1}b \quad\text{and}\quad k = c - h^{\mathrm{T}}A h = c - b^{\mathrm{T}} (A+A^{\mathrm{T}})^{-1} A (A+A^{\mathrm{T}})^{-1}b</math>.
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| ==Relation to the graph==
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| [[Image:H shift.png|thumb|right|250px|Graphs of quadratic functions shifted to the right by ''h'' = 0, 5, 10, and 15.]]
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| [[Image:V shift.png|thumb|right|250px|Graphs of quadratic functions shifted upward by ''k'' = 0, 5, 10, and 15.]]
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| [[Image:HV shift.png|thumb|right|250px|Graphs of quadratic functions shifted upward and to the right by 0, 5, 10, and 15.]]
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| In [[analytic geometry]], the graph of any [[quadratic function]] is a [[parabola]] in the ''xy''-plane. Given a quadratic polynomial of the form
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| :<math>(x-h)^2 + k \quad\text{or}\quad a(x-h)^2 + k</math>
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| the numbers ''h'' and ''k'' may be interpreted as the [[Cartesian coordinates]] of the vertex of the parabola. That is, ''h'' is the ''x''-coordinate of the axis of symmetry, and ''k'' is the [[maxima and minima|minimum value]] (or maximum value, if ''a'' < 0) of the quadratic function.
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| In other words, the graph of the function ''ƒ''(''x'') = ''x''<sup>2</sup> is a parabola whose vertex is at the origin (0, 0). Therefore, the graph of the function ''ƒ''(''x'' − ''h'') = (''x'' − ''h'')<sup>2</sup> is a parabola shifted to the right by ''h'' whose vertex is at (''h'', 0), as shown in the top figure. In contrast, the graph of the function ''ƒ''(''x'') + ''k'' = ''x''<sup>2</sup> + ''k'' is a parabola shifted upward by ''k'' whose vertex is at (0, ''k''), as shown in the center figure. Combining both horizontal and vertical shifts yields ''ƒ''(''x'' − ''h'') + ''k'' = (''x'' − ''h'')<sup>2</sup> + ''k'' is a parabola shifted to the right by ''h'' and upward by ''k'' whose vertex is at (''h'', ''k''), as shown in the bottom figure.
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| ==Solving quadratic equations==
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| Completing the square may be used to solve any [[Quadratic equation#By completing the square|quadratic equation]]. For example:
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| :<math>x^2 + 6x + 5 = 0,\,\!</math>
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| The first step is to complete the square:
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| :<math>(x+3)^2 - 4 = 0.\,\!</math>
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| Next we solve for the squared term:
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| :<math>(x+3)^2 = 4.\,\!</math>
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| Then either
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| :<math>x+3 = -2 \quad\text{or}\quad x+3 = 2,</math>
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| and therefore
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| :<math>x = -5 \quad\text{or}\quad x = -1.</math>
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| This can be applied to any quadratic equation. When the ''x''<sup>2</sup> has a coefficient other than 1, the first step is to divide out the equation by this coefficient: for an example see the non-monic case below.
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| ===Irrational and complex roots===
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| Unlike methods involving [[factorization|factoring]] the equation, which is only reliable if the roots are [[Rational number|rational]], completing the square will find the roots of a quadratic equation even when those roots are [[irrational number|irrational]] or [[Complex number|complex]]. For example, consider the equation
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| :<math>x^2 - 10x + 18 = 0.\,\!</math>
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| Completing the square gives
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| :<math>(x-5)^2 - 7 = 0,\,\!</math>
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| so
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| :<math>(x-5)^2 = 7.\,\!</math>
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| Then either
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| :<math>x-5 = -\sqrt{7} \quad\text{or}\quad x-5 = \sqrt{7},\,</math>
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| so
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| : <math> x = 5 - \sqrt{7}\quad\text{or}\quad x = 5 + \sqrt{7}. \, </math>
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| In terser language:
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| :<math>x = 5 \pm \sqrt{7}.\,</math>
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| Equations with complex roots can be handled in the same way. For example:
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| :<math>\begin{array}{c}
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| x^2 + 4x + 5 \,=\, 0 \\[6pt]
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| (x+2)^2 + 1 \,=\, 0 \\[6pt]
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| (x+2)^2 \,=\, -1 \\[6pt]
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| x+2 \,=\, \pm i \\[6pt]
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| x \,=\, -2 \pm i.
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| \end{array}
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| </math>
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| ===Non-monic case===
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| For an equation involving a non-monic quadratic, the first step to solving them is to divide through by the coefficient of ''x''<sup>2</sup>. For example:
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| :<math>\begin{array}{c}
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| 2x^2 + 7x + 6 \,=\, 0 \\[6pt]
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| x^2 + \tfrac{7}{2}x + 3 \,=\, 0 \\[6pt]
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| \left(x+\tfrac{7}{4}\right)^2 - \tfrac{1}{16} \,=\, 0 \\[6pt]
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| \left(x+\tfrac{7}{4}\right)^2 \,=\, \tfrac{1}{16} \\[6pt]
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| x+\tfrac{7}{4} = \tfrac{1}{4} \quad\text{or}\quad x+\tfrac{7}{4} = -\tfrac{1}{4} \\[6pt]
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| x = -\tfrac{3}{2} \quad\text{or}\quad x = -2.
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| \end{array}
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| </math>
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| ==Other applications==
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| ===Integration===
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| Completing the square may be used to evaluate any integral of the form
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| :<math>\int\frac{dx}{ax^2+bx+c}</math>
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| using the basic integrals
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| :<math>\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| +C \quad\text{and}\quad
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| \int\frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) +C.</math>
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| For example, consider the integral
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| :<math>\int\frac{dx}{x^2 + 6x + 13}.</math>
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| Completing the square in the denominator gives:
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| :<math>\int\frac{dx}{(x+3)^2 + 4} \,=\, \int\frac{dx}{(x+3)^2 + 2^2}.</math>
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| This can now be evaluated by using the [[integration by substitution|substitution]]
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| ''u'' = ''x'' + 3, which yields
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| :<math>\int\frac{dx}{(x+3)^2 + 4} \,=\, \frac{1}{2}\arctan\left(\frac{x+3}{2}\right)+C.</math>
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| ===Complex numbers===
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| Consider the expression
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| :<math> |z|^2 - b^*z - bz^* + c,\,</math>
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| where ''z'' and ''b'' are [[complex number]]s, ''z''<sup>*</sup> and ''b''<sup>*</sup> are the [[complex conjugate]]s of ''z'' and ''b'', respectively, and ''c'' is a [[real number]]. Using the identity |''u''|<sup>2</sup> = ''uu''<sup>*</sup> we can rewrite this as
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| :<math> |z-b|^2 - |b|^2 + c , \,\!</math>
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| which is clearly a real quantity. This is because
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| :<math>
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| \begin{align}
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| |z-b|^2 &{}= (z-b)(z-b)^*\\
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| &{}= (z-b)(z^*-b^*)\\
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| &{}= zz^* - zb^* - bz^* + bb^*\\
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| &{}= |z|^2 - zb^* - bz^* + |b|^2 .
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| \end{align}</math>
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| As another example, the expression
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| :<math> ax^2 + by^2 + c , \,\!</math>
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| where ''a'', ''b'', ''c'', ''x'', and ''y'' are real numbers, with ''a'' > 0 and ''b'' > 0, may be expressed in terms of the square of the [[absolute value]] of a complex number. Define
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| :<math> z = \sqrt{a}\,x + i \sqrt{b} \,y . </math>
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| Then
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| :<math> | |
| \begin{align}
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| |z|^2 &{}= z z^*\\
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| &{}= (\sqrt{a}\,x + i \sqrt{b}\,y)(\sqrt{a}\,x - i \sqrt{b}\,y) \\
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| &{}= ax^2 - i\sqrt{ab}\,xy + i\sqrt{ba}\,yx - i^2by^2 \\
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| &{}= ax^2 + by^2 ,
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| \end{align}</math>
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| so
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| :<math> ax^2 + by^2 + c = |z|^2 + c . \,\!</math>
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| ==Geometric perspective==
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| [[Image:Completing the square 307.PNG|right]]
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| Consider completing the square for the equation
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| :<math>x^2 + bx = a.\,</math>
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| Since ''x''<sup>2</sup> represents the area of a square with side of length ''x'', and ''bx'' represents the area of a rectangle with sides ''b'' and ''x'', the process of completing the square can be viewed as visual manipulation of rectangles.
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| Simple attempts to combine the ''x''<sup>2</sup> and the ''bx'' rectangles into a larger square result in a missing corner. The term (''b''/2)<sup>2</sup> added to each side of the above equation is precisely the area of the missing corner, whence derives the terminology "completing the square". [http://maze5.net/?page_id=467]
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| ==A variation on the technique==
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| As conventionally taught, completing the square consists of adding the third term, ''v''<sup> 2</sup> to
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| :<math>u^2 + 2uv\,</math>
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| to get a square. There are also cases in which one can add the middle term, either 2''uv'' or −2''uv'', to
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| :<math>u^2 + v^2\,</math>
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| to get a square.
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| ===Example: the sum of a positive number and its reciprocal===
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| By writing
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| :<math>
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| \begin{align}
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| x + {1 \over x} &{} = \left(x - 2 + {1 \over x}\right) + 2\\
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| &{}= \left(\sqrt{x} - {1 \over \sqrt{x}}\right)^2 + 2
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| \end{align}</math>
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| we show that the sum of a positive number ''x'' and its reciprocal is always greater than or equal to 2. The square of a real expression is always greater than or equal to zero, which gives the stated bound; and here we achieve 2 just when ''x'' is 1, causing the square to vanish.
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| ===Example: factoring a simple quartic polynomial===
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| Consider the problem of factoring the polynomial
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| :<math>x^4 + 324 . \,\!</math>
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| This is
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| :<math>(x^2)^2 + (18)^2, \,\!</math>
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| so the middle term is 2(''x''<sup>2</sup>)(18) = 36''x''<sup>2</sup>. Thus we get
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| :<math>\begin{align} x^4 + 324 &{}= (x^4 + 36x^2 + 324 ) - 36x^2 \\
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| &{}= (x^2 + 18)^2 - (6x)^2 =\text{a difference of two squares} \\
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| &{}= (x^2 + 18 + 6x)(x^2 + 18 - 6x) \\
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| &{}= (x^2 + 6x + 18)(x^2 - 6x + 18)
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| \end{align}</math>
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| (the last line being added merely to follow the convention of decreasing degrees of terms).
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| ==References==
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| {{reflist}}
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| *Algebra 1, Glencoe, ISBN 0-07-825083-8, pages 539–544
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| *Algebra 2, Saxon, ISBN 0-939798-62-X, pages 214–214, 241–242, 256–257, 398–401
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| ==External links==
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| *{{planetmath reference|id=4237|title=Completing the square}}
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| *[http://education-portal.com/academy/lesson/how-to-complete-the-square.html How to Complete the Square, Education Portal Academy]
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| [[Category:Elementary algebra]]
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| [[Category:Articles containing proofs]]
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| [[ja:二次方程式#平方完成]]
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