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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]])]]
| | == 第一千五百四十章の生き物 == |
|
| |
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| In [[elementary algebra]], '''completing the square''' is a technique for converting a [[quadratic polynomial]] of the form
| | 転送のためにオプトインした後、その上に黒の服と黒い石のベンチにとどまるすべての若者を与える [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-7.html カシオソーラー時計]。<br>見て、多くの目には<br>、シャオヤンがゆっくりと笑みを浮かべて立ち上がって、ショックを受けて身長、光っぼかし、しかしその影、それはまた、上記の空に現れ、クールで明るく笑い、「スイング」を通過させ、オープン [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-10.html カシオ 腕時計 スタンダード]。<br><br>「仮想児童前任者がDanta魂盗む芸術についてはあまり聞いていた、天府は今Dantaはメンバーであり、天府の国民として、シャオヤン魂が、いくつかの仮想サブ前任者がアドバイスを！まだ翼を願っ世代Dantaたい触れる！ [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-9.html カシオ 掛け時計] '<br>ヤンの<br>第一千五百四十章の生き物<br>ヤンの<br>第一千五百四十章の生き物<br>シャオヤンロング笑い声が聞こえ<br>、誰もが密かにある恥の多くは、これらの言葉が、非常に良いではない、ああ、それは家族の魂と天府提携だが、それは本当に深い恨みです<br>これと他の資格のない、 [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-2.html カシオ腕時計 g-shock] 'その後、私のアドバイスにかみそりを、も、あなたの先生が「<br>医学'ちり |
| | 相关的主题文章： |
| | <ul> |
| | |
| | <li>[http://www.91taoren.com/home.php?mod=space&uid=11104 http://www.91taoren.com/home.php?mod=space&uid=11104]</li> |
| | |
| | <li>[http://www.standardall.com/plus/feedback.php?aid=8996 http://www.standardall.com/plus/feedback.php?aid=8996]</li> |
| | |
| | <li>[http://www.lcgdz.com/plus/feedback.php?aid=40 http://www.lcgdz.com/plus/feedback.php?aid=40]</li> |
| | |
| | </ul> |
|
| |
|
| :<math>ax^2 + bx + c\,\!</math>
| | == 「強制的に彼によって連れ去ら == |
|
| |
|
| to the form
| | ヘッド、Xiupao曹操Yingさん、風の前の3ストロークはトリオはすぐにその後流さスターフィールドの嵐に対する出口方向を育てた、とされている [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-1.html カシオ 腕時計 チタン]。<br><br>「教師、シャオヤン！ [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-12.html カシオ 腕時計 ソーラー] '<br><br>を強制的に奪われた場合には、曹操Yingさん急いで [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-13.html カシオ腕時計 メンズ]。<br><br>玄コングサブしっかりシャオヤンを見て目がそれらの炎の炎の中で周囲のZiheiに「色」を生きてきた、眉をひそめ、彼に急いではない、一般的な意識を持っているようだが、この中高い温度のようなものを作るために十分に長い時間に最適な場所は、バケット銅像に強いホールドを持っている [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-6.html casio 腕時計 メンズ]。<br><br>「強制的に彼によって連れ去ら。 [http://alleganycountyfair.org/sitemap.xml http://alleganycountyfair.org/sitemap.xml] '<br><br>心は一瞬ためらった、この状態からシャオヤンのブレークは、いくつかのダメージを与えますが、子供は最終的に玄コングは、一口ですが、シャオ·円高への一方向のものが存在しない、あまりにも長い我慢できないこれと他の高温。<br><br>この道を通るという考えを気にし、移動の玄コングサブ体格、すなわち |
| | 相关的主题文章： |
| | <ul> |
| | |
| | <li>[http://www.cap-mould.com/plus/view.php?aid=75648 http://www.cap-mould.com/plus/view.php?aid=75648]</li> |
| | |
| | <li>[http://tjcutao.com/discuzx31gbk/home.php?mod=space&uid=41211 http://tjcutao.com/discuzx31gbk/home.php?mod=space&uid=41211]</li> |
| | |
| | <li>[http://www.immobilien-schacht.de/index.php?site=guestbook&action=add http://www.immobilien-schacht.de/index.php?site=guestbook&action=add]</li> |
| | |
| | </ul> |
|
| |
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| : <math> a(\cdots\cdots)^2 + \mbox{constant}.\, </math>
| | == 」シャオヤンは微笑んで、すぐに道をうなずいた == |
|
| |
|
| In this context, "constant" means not depending on ''x''. The expression inside the parenthesis is of the form (''x'' + constant). Thus
| | 千 [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-4.html カシオ 腕時計 バンド] '医学'スクエア女将ポイント安く、幸いにも、その前の密かリマインダー、シャオヤンそのまだかなり良い印象を占め、彼女は問題安くポイントを占めていませんしましょう。<br><br>薬「八尾広場、メインなしの意見は、私は三味線を取った場合は、「」材料を受け入れるように [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-13.html casio 腕時計 edifice]。」シャオヤンは微笑んで、すぐに道をうなずいた [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-8.html casio 腕時計 スタンダード]。<br>聞い<br>、八尾広場、メイン頬が突然エクスタシーのヒントを浮上、ギャングゆうチー山の片側が激しいポンドのテーブルであり、うなずいた、ニュースキン盛は言った： [http://nrcil.net/sitemap.xml http://nrcil.net/sitemap.xml] 'のように！'<br><br>談話はチーシャン八尾広場中断された、怒りのタッチで抑圧された音声の中に沈んだ少しメインアイ「カラー」、Meimouステアリングチー山は、「古い、皇室の側室では、黒の人は頭を持っている尊重しましょうだから珍しい尊重しますが、立ち入りあなたの次の順序のための私に千 '薬'広場を得ることができないでください！ [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-4.html カシオ 腕時計 バンド] '<br>千帝都のこのような急成長事業のブラックスクエアが可能な「薬」だけでなく、これと他の旧ヤンリアン」の強さを<br> |
| :<math>ax^2 + bx + c\,\!</math> is converted to | | 相关的主题文章： |
| | <ul> |
| | |
| | <li>[http://www.cclchinese.com/?action-viewcomment-type-news-itemid-8008 http://www.cclchinese.com/?action-viewcomment-type-news-itemid-8008]</li> |
| | |
| | <li>[http://www.topledtube.com/plus/feedback.php?aid=150 http://www.topledtube.com/plus/feedback.php?aid=150]</li> |
| | |
| | <li>[http://www.chantal.cn/bbs/forum.php?mod=viewthread&tid=58642&fromuid=24048 http://www.chantal.cn/bbs/forum.php?mod=viewthread&tid=58642&fromuid=24048]</li> |
| | |
| | </ul> |
|
| |
|
| : <math> a(x + h)^2 + k\, </math>
| | == 気晴らし後の魂のカストディアン、より強力な == |
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| for some values of ''h'' and ''k''.
| | 気晴らし後の魂のカストディアン、より強力な [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-2.html casio 腕時計 デジタル]<br><br>3「カラー」仏蓮の怒り、シャオヤンカード、現在、彼の強さで、一度だけ、その後にキャストすることができますが、デ力の状態に入ることを許可されている、つまり、彼は唯一のチャンスである、ハスを解雇されていない場合彼は古い悪魔の魔法のための打撃は致命的な [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-15.html カシオ ソーラー 腕時計] 'セックス'であることを確認する必要がありますので、その後、今日は最悪の状況があるかもしれません、殺すために古い悪魔の魔法！<br><br>「プチ！ [http://www.ispsc.edu.ph/nav/japandi/casio-rakuten-8.html カシオ 掛け時計] '<br>小型医療セントの魔法でその古い悪魔の対立の間<br>シャオヤンの心が点滅するだけでなく、血液スプレーは横に振った、現時点では、彼女はより多くの淡い頬盛は、戦闘がレベルだった間に、各レベルには、彼女の完全な二、三のレベルよりも高い、古い悪魔の魔法はもちろんのこと、ギャップを通過することは極めて困難である<br><br>古い悪魔の魔法を見てニヤリ顔を見て、再び嘔吐小型医療セントを見て、シャオヤンは次第に、肌、黒い瞳を横に振った |
| | | 相关的主题文章： |
| Completing the square is used in
| | <ul> |
| * solving [[quadratic equation]]s,
| | |
| * graphing [[quadratic function]]s,
| | <li>[http://www.chcmp.com/bbs/forum.php?mod=viewthread&tid=136767 http://www.chcmp.com/bbs/forum.php?mod=viewthread&tid=136767]</li> |
| * evaluating [[integral]]s in calculus, such as [[Gaussian Integral|Gaussian integrals]] with a linear term in the exponent
| | |
| * finding [[Laplace transforms]].
| | <li>[http://bbs.shenghuoqu.com.cn/forum.php?mod=viewthread&tid=1303306&fromuid=119459 http://bbs.shenghuoqu.com.cn/forum.php?mod=viewthread&tid=1303306&fromuid=119459]</li> |
| | | |
| 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]].
| | <li>[http://www.kuts.cn/dz2/forum.php?mod=viewthread&tid=43138 http://www.kuts.cn/dz2/forum.php?mod=viewthread&tid=43138]</li> |
| | | |
| ==Overview==
| | </ul> |
| | |
| ===Background===
| |
| There is a simple formula in [[elementary algebra]] for computing the [[square (algebra)|square]] of a [[binomial]]:
| |
| | |
| :<math>(x + p)^2 \,=\, x^2 + 2px + p^2.\,\!</math>
| |
| | |
| For example:
| |
| | |
| :<math>\begin{alignat}{2}
| |
| (x+3)^2 \,&=\, x^2 + 6x + 9 && (p=3)\\[3pt]
| |
| (x-5)^2 \,&=\, x^2 - 10x + 25\qquad && (p=-5).
| |
| \end{alignat}
| |
| </math>
| |
| | |
| 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>.
| |
| | |
| ===Basic example===
| |
| Consider the following quadratic [[polynomial]]:
| |
| | |
| :<math>x^2 + 10x + 28.\,\!</math>
| |
| | |
| This quadratic is not a perfect square, since 28 is not the square of 5:
| |
| | |
| :<math>(x+5)^2 \,=\, x^2 + 10x + 25.\,\!</math>
| |
| | |
| However, it is possible to write the original quadratic as the sum of this square and a constant:
| |
| | |
| :<math>x^2 + 10x + 28 \,=\, (x+5)^2 + 3.</math>
| |
| | |
| This is called '''completing the square'''.
| |
| | |
| ===General description===
| |
| Given any [[Monic polynomial|monic]] quadratic
| |
| | |
| :<math>x^2 + bx + c,\,\!</math>
| |
| | |
| it is possible to form a square that has the same first two terms:
| |
| | |
| :<math>\left(x+\tfrac{1}{2} b\right)^2 \,=\, x^2 + bx + \tfrac{1}{4}b^2.</math>
| |
| | |
| This square differs from the original quadratic only in the value of the constant
| |
| term. Therefore, we can write
| |
| | |
| :<math>x^2 + bx + c \,=\, \left(x + \tfrac{1}{2}b\right)^2 + k,</math>
| |
| | |
| where ''k'' is a constant. This operation is known as '''completing the square'''.
| |
| For example:
| |
| | |
| :<math>\begin{alignat}{1}
| |
| x^2 + 6x + 11 \,&=\, (x+3)^2 + 2 \\[3pt]
| |
| x^2 + 14x + 30 \,&=\, (x+7)^2 - 19 \\[3pt]
| |
| x^2 - 2x + 7 \,&=\, (x-1)^2 + 6.
| |
| \end{alignat}
| |
| </math>
| |
| | |
| ===Non-monic case===
| |
| Given a quadratic polynomial of the form
| |
| :<math>ax^2 + bx + c\,\!</math>
| |
| it is possible to factor out the coefficient ''a'', and then complete the square for the resulting [[monic polynomial]].
| |
| | |
| Example:
| |
| :<math>
| |
| \begin{align}
| |
| 3x^2 + 12x + 27 &= 3(x^2+4x+9)\\
| |
| &{}= 3\left((x+2)^2 + 5\right)\\
| |
| &{}= 3(x+2)^2 + 15
| |
| \end{align}</math>
| |
| This allows us to write any quadratic polynomial in the form
| |
| :<math>a(x-h)^2 + k.\,\!</math>
| |
| | |
| ===Formula===
| |
| The result of completing the square may be written as a formula. For the general case:<ref>{{cite book
| |
| |title=Precalculus: Building Concepts and Connections
| |
| |first1=Revathi
| |
| |last1=Narasimhan
| |
| |publisher=Cengage Learning
| |
| |year=2008
| |
| |isbn=0-618-41301-4
| |
| |pages=133–134
| |
| |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]
| |
| </ref>
| |
| | |
| :<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>
| |
| | |
| Specifically, when ''a''=1:
| |
| | |
| :<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>
| |
| | |
| The matrix case looks very similar:
| |
| | |
| :<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>
| |
| | |
| where <math>A</math> has to be symmetric.
| |
| | |
| If <math>A</math> is not symmetric the formulae for <math>h</math> and <math>k</math> have
| |
| to be generalized to:
| |
| | |
| :<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>.
| |
| | |
| ==Relation to the graph==
| |
| [[Image:H shift.png|thumb|right|250px|Graphs of quadratic functions shifted to the right by ''h'' = 0, 5, 10, and 15.]]
| |
| [[Image:V shift.png|thumb|right|250px|Graphs of quadratic functions shifted upward by ''k'' = 0, 5, 10, and 15.]]
| |
| [[Image:HV shift.png|thumb|right|250px|Graphs of quadratic functions shifted upward and to the right by 0, 5, 10, and 15.]]
| |
| | |
| In [[analytic geometry]], the graph of any [[quadratic function]] is a [[parabola]] in the ''xy''-plane. Given a quadratic polynomial of the form
| |
| | |
| :<math>(x-h)^2 + k \quad\text{or}\quad a(x-h)^2 + k</math>
| |
| | |
| 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.
| |
| | |
| 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.
| |
| | |
| ==Solving quadratic equations==
| |
| Completing the square may be used to solve any [[Quadratic equation#By completing the square|quadratic equation]]. For example:
| |
| | |
| :<math>x^2 + 6x + 5 = 0,\,\!</math>
| |
| | |
| The first step is to complete the square:
| |
| | |
| :<math>(x+3)^2 - 4 = 0.\,\!</math>
| |
| | |
| Next we solve for the squared term:
| |
| | |
| :<math>(x+3)^2 = 4.\,\!</math>
| |
| | |
| Then either
| |
| | |
| :<math>x+3 = -2 \quad\text{or}\quad x+3 = 2,</math>
| |
| | |
| and therefore
| |
| | |
| :<math>x = -5 \quad\text{or}\quad x = -1.</math> | |
| | |
| 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.
| |
| | |
| ===Irrational and complex roots===
| |
| 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
| |
| | |
| :<math>x^2 - 10x + 18 = 0.\,\!</math>
| |
| | |
| Completing the square gives
| |
| | |
| :<math>(x-5)^2 - 7 = 0,\,\!</math>
| |
| | |
| so
| |
| | |
| :<math>(x-5)^2 = 7.\,\!</math>
| |
| | |
| Then either
| |
| | |
| :<math>x-5 = -\sqrt{7} \quad\text{or}\quad x-5 = \sqrt{7},\,</math>
| |
| | |
| so
| |
| | |
| : <math> x = 5 - \sqrt{7}\quad\text{or}\quad x = 5 + \sqrt{7}. \, </math> | |
| | |
| In terser language:
| |
| | |
| :<math>x = 5 \pm \sqrt{7}.\,</math>
| |
| | |
| Equations with complex roots can be handled in the same way. For example:
| |
| | |
| :<math>\begin{array}{c}
| |
| x^2 + 4x + 5 \,=\, 0 \\[6pt]
| |
| (x+2)^2 + 1 \,=\, 0 \\[6pt]
| |
| (x+2)^2 \,=\, -1 \\[6pt]
| |
| x+2 \,=\, \pm i \\[6pt]
| |
| x \,=\, -2 \pm i.
| |
| \end{array}
| |
| </math>
| |
| | |
| ===Non-monic case===
| |
| 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:
| |
| | |
| :<math>\begin{array}{c}
| |
| 2x^2 + 7x + 6 \,=\, 0 \\[6pt]
| |
| x^2 + \tfrac{7}{2}x + 3 \,=\, 0 \\[6pt]
| |
| \left(x+\tfrac{7}{4}\right)^2 - \tfrac{1}{16} \,=\, 0 \\[6pt]
| |
| \left(x+\tfrac{7}{4}\right)^2 \,=\, \tfrac{1}{16} \\[6pt]
| |
| x+\tfrac{7}{4} = \tfrac{1}{4} \quad\text{or}\quad x+\tfrac{7}{4} = -\tfrac{1}{4} \\[6pt]
| |
| x = -\tfrac{3}{2} \quad\text{or}\quad x = -2.
| |
| \end{array}
| |
| </math> | |
| | |
| ==Other applications==
| |
| | |
| ===Integration===
| |
| Completing the square may be used to evaluate any integral of the form
| |
| | |
| :<math>\int\frac{dx}{ax^2+bx+c}</math>
| |
| | |
| using the basic integrals
| |
| | |
| :<math>\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| +C \quad\text{and}\quad | |
| \int\frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) +C.</math>
| |
| | |
| For example, consider the integral
| |
| | |
| :<math>\int\frac{dx}{x^2 + 6x + 13}.</math>
| |
| | |
| Completing the square in the denominator gives:
| |
| | |
| :<math>\int\frac{dx}{(x+3)^2 + 4} \,=\, \int\frac{dx}{(x+3)^2 + 2^2}.</math>
| |
| | |
| 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>
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| \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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