利用者:ShuBraque/sandbox/二次方程式の解の公式

2次方程式の解の公式

en:代数学基礎において、2次方程式の解の公式とは2次方程式の解を求める式である。因数分解や平方完成、en:グラフを書くなど解の公式を使う以外にも2次方程式を解くためには方法がある。しかし、2次方程式の解の公式を使うことが最も便利な方法であることがしばしばある。

2次方程式の一般形は

$ax^{2}+bx+c=0.$

ここで x は未知数を示しており、 a 、 b 、 c は定数である（ただし a は0ではない）。2次方程式が2次方程式の解の公式を満たすことを、2次方程式の解の公式を2次方程式に代入することで確かめることができる。 2次方程式の解の公式によって与えられるそれぞれの解はen:根と呼ばれる。

解の公式の導出

平方完成を理解していれば、2次方程式の解の公式を導出することができる。このため、2次方程式の解の公式の導出は学生に課題として与えられることがあり、この課題を行うことで学生はこの重要な公式を再発見することが可能なのである。陽関数表示の導出は次の通り。

a が0でないことから、 a で割ることが可能である。2次方程式を a で割る。

$x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0.$

c/a を等式の両辺から引く。すると次のようになる。

$x^{2}+{\frac {b}{a}}x=-{\frac {c}{a}}.$

この2次方程式は平方完成が適用可能な形となっている。よって、等式の両辺に定数を足し、等式の左辺を平方完成とする。

$x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}=-{\frac {c}{a}}+\left({\frac {b}{2a}}\right)^{2},$

これを変形する。

$\left(x+{\frac {b}{2a}}\right)^{2}=-{\frac {c}{a}}+{\frac {b^{2}}{4a^{2}}}$.

最後に、右辺の項を変形し公分母を得ることで、次の式を得る。

$\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}.$

等式が平方完成された。等式の両辺の平方根を取ることで次の式を得る。

$x+{\frac {b}{2a}}=\pm {\frac {\sqrt {b^{2}-4ac\ }}{2a}}.$

x イコールの形に直すことで2次方程式の解の公式を得る。

$x={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}.$

このプラスマイナス記号 "±" は次の2つを示している。

$x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\quad {\text{and}}\quad x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}$

これらは2次方程式の解である。この導出以外にも様々な導出方法があり、それぞれ多少の違いがあるが、ほとんどのものが$a$の操作に関するものである。

一部の文献、特に古いものでは $ax^{2}-2bx+c=0$ や $ax^{2}+2bx+c=0$のような異なるパラメータ表示をしていることがあるwhere b has a magnitude one half of the more common one. These result in slightly different forms for the solution, but are otherwise equivalent.

発展の歴史

ラファエロのアテナイの学堂の中のユークリッド。

二次方程式に解を与える最初期の方法は幾何学的であった。バビロニアのくさび文字で書かれた文字板には二次方程式を解くことに単純化可能な問題が含まれていた。エジプト中王国の時代 (2050 BC to 1650 BC) にまで遡る、エジプトのen:Berlin Papyrusには二項の二次方程式の解が含まれていた。

ギリシャの数学者ユークリッド (およそ 300 BC) は原論という自身の著作のなかで二次方程式を解くのに幾何学的方法を使った。原論は非常に大きな影響を与えた数学の学術文献である。およそ200 BCの中国の九章算術には二次方程式に対する解法が登場する。ギリシャの数学者ディオファントス (およそ 250 BC)は、自身の著作算術において二次方程式を解いたが、彼の手法はユークリッドの幾何学的手法と比較してより代数学的であったとされる。ディオファントスの解は、たとえ2つの根が共に正であってもひとつの根のみを与える。

インドの数学者であるブラーマグプタ (597–668 AD)は自身の学術論文en:Brāhmasphuṭasiddhāntaの中で明示的に二次方程式の解の公式をを示した。Brāhmasphuṭasiddhāntaは628 ADに出版されたが、記号（symbol）ではなく言葉を使って書かれていた。ブラーマグプタの二次方程式 $ax^{2}+bx=c$ の解は次の通りである。"To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value."これは次に等しい。

$x={\frac {{\sqrt {4ac+b^{2}}}-b}{2a}}.$

初期のギリシャおよびインドの数学者に影響を受けた9世紀のペルシャの数学者フワーリズミーは、二次方程式を代数学的に解いた。すべてのケースに対して有効な二次方程式の解の公式は1594年にシモン・ステヴィンによって最初に得られた。1637年にはルネ・デカルトによってen:La Géométrieが出版されたが、この本には今日私たちが知っている形式で二次方程式の解の公式が収録されている。一般解が現代的な数学の学術文献に初めて登場したのは1896年で、Henry Heatonによる論文のなかで言及されたものである

Importance of this solution

Among the many equations that one encounters while studying algebra, the quadratic formula is one of the most important, and is considered the most useful method of solving quadratic equations. Unlike some other solution methods such as factoring, the quadratic formula can be used to solve any quadratic equation. Many equations that do not initially appear to be quadratic can be put into quadratic form, and solved using the quadratic formula. For these reasons, it is often memorized.

Completing the square also allows for the solution of all quadratics, as it is mathematically equivalent, but the quadratic formula gives a result without the need for so much algebraic manipulation. As such, it is generally considered more practical to use the formula. Completing the square is very useful for other purposes, such as putting the equations for conic sections into standard form.

Other derivations

A number of alternative derivations of the quadratic formula can be found in the literature. These derivations either (a) are simpler than the standard completing the square method, (b) represent interesting applications of other frequently used techniques in algebra, or (c) offer insight into other areas of mathematics.

Alternate method of completing the square

The great majority of algebra texts published over the last several decades teach completing the square using the sequence presented earlier: (1) divide each side by a, (2) rearrange, (3) then add the square of one-half of b/a.

As pointed out by Larry Hoehn in 1975, completing the square can be accomplished by a different sequence that leads to a simpler sequence of intermediate terms: (1) multiply each side by 4a, (2) rearrange, (3) then add $b^{2}$.

In other words, the quadratic formula can be derived as follows:

${\begin{aligned}ax^{2}+bx+c&=0\\4a^{2}x^{2}+4abx+4ac&=0\\4a^{2}x^{2}+4abx&=-4ac\\4a^{2}x^{2}+4abx+b^{2}&=b^{2}-4ac\\(2ax+b)^{2}&=b^{2}-4ac\\2ax+b&=\pm {\sqrt {b^{2}-4ac}}\\2ax&=-b\pm {\sqrt {b^{2}-4ac}}\\x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\\\end{aligned}}$

This actually represents an ancient derivation of the quadratic formula, and was known to the Hindus at least as far back as 1025 AD. Compared with the derivation in standard usage, this alternate derivation is shorter, involves fewer computations with literal coefficients, avoids fractions until the last step, has simpler expressions, and uses simpler math. As Hoehn states, "it is easier 'to add the square of b' than it is 'to add the square of half the coefficient of the x term'".

By substitution

Another technique is solution by substitution. In this technique, we substitute $x=y+m$ into the quadratic to get:

$a(y+m)^{2}+b(y+m)+c=0$

Expanding the result and then collecting the powers of $y$ produces:

$ay^{2}+y(2am+b)+(am^{2}+bm+c)=0$

We have not yet imposed a second condition on $y$ and $m$, so we now choose m so that the middle term vanishes. That is, $2am+b=0$ or $m={\frac {-b}{2a}}$. Subtracting the constant term from both sides of the equation (to move it to the right hand side) and then dividing by a gives:

$y^{2}={\frac {-(am^{2}+bm+c)}{a}}$

Substituting for $m$ gives:

$y^{2}={\frac {-({\frac {b^{2}}{4a}}+{\frac {-b^{2}}{2a}}+c)}{a}}={\frac {b^{2}-4ac}{4a^{2}}}$

Therefore $y=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}$; substituting $x=y+m=y-{\frac {b}{2a}}$ provides the quadratic formula.

By using algebraic identities

Let the roots of the standard quadratic equation be $r_{1}$ and $r_{2}$. At this point, we recall the identity:

$(r_{1}-r_{2})^{2}=(r_{1}+r_{2})^{2}-4r_{1}r_{2}$

Taking square root on both sides, we get

$r_{1}-r_{2}=\pm {\sqrt {(r_{1}+r_{2})^{2}-4r_{1}r_{2}}}$

Since the coefficient a ≠ 0, we can divide the standard equation by a to obtain a quadratic polynomial having the same roots. Namely,

$x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=(x-r_{1})(x-r_{2})=x^{2}-(r_{1}+r_{2})x+r_{1}r_{2}.$

From this we can see that the sum of the roots of the standard quadratic equation is given by $-{\frac {b}{a}}$, and the product of those roots is given by ${\frac {c}{a}}.$

Hence the identity can be rewritten as:

$r_{1}-r_{2}=\pm {\sqrt {(-{\frac {b}{a}})^{2}-4{\frac {c}{a}}}}=\pm {\sqrt {{\frac {b^{2}}{a^{2}}}-{\frac {4ac}{a^{2}}}}}=\pm {\frac {\sqrt {b^{2}-4ac}}{a}}$

Now,

$r_{1}={\frac {(r_{1}+r_{2})+(r_{1}-r_{2})}{2}}={\frac {-{\frac {b}{a}}\pm {\frac {\sqrt {b^{2}-4ac}}{a}}}{2}}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.$

Since, $r_{2}=-r_{1}-{\frac {b}{a}}$, if we take $r_{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}$ then we obtain $r_{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}$ and if we instead take $r_{1}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}$ then we calculate that $r_{2}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}.$ Combining these results by using the standard shorthand, we have that the solutions of the quadratic equation are given by:

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.$

By Lagrange resolvents

詳細は「Lagrange resolvents」を参照

An alternative way of deriving the quadratic formula is via the method of Lagrange resolvents, which is an early part of Galois theory. This method can be generalized to give the roots of cubic polynomials and quartic polynomials, and leads to Galois theory, which allows one to understand the solution of algebraic equations of any degree in terms of the symmetry group of their roots, the Galois group.

This approach focuses on the roots more than on rearranging the original equation. Given a monic quadratic polynomial

$x^{2}+px+q,$

assume that it factors as

$x^{2}+px+q=(x-\alpha )(x-\beta ),$

Expanding yields

$x^{2}+px+q=x^{2}-(\alpha +\beta )x+\alpha \beta ,$

where $p=-(\alpha +\beta )$ and $q=\alpha \beta$.

Since the order of multiplication does not matter, one can switch $\alpha$ and $\beta$ and the values of p and q will not change: one says that p and q are symmetric polynomials in $\alpha$ and $\beta$. In fact, they are the elementary symmetric polynomials – any symmetric polynomial in $\alpha$ and $\beta$ can be expressed in terms of $\alpha +\beta$ and $\alpha \beta .$ The Galois theory approach to analyzing and solving polynomials is: given the coefficients of a polynomial, which are symmetric functions in the roots, can one "break the symmetry" and recover the roots? Thus solving a polynomial of degree n is related to the ways of rearranging ("permuting") n terms, which is called the symmetric group on n letters, and denoted $S_{n}.$ For the quadratic polynomial, the only way to rearrange two terms is to swap them ("transpose" them), and thus solving a quadratic polynomial is simple.

To find the roots $\alpha$ and $\beta ,$ consider their sum and difference:

${\begin{aligned}r_{1}&=\alpha +\beta \\r_{2}&=\alpha -\beta .\end{aligned}}$

These are called the Lagrange resolvents of the polynomial; notice that one of these depends on the order of the roots, which is the key point. One can recover the roots from the resolvents by inverting the above equations:

${\begin{aligned}\alpha &=\textstyle {\frac {1}{2}}\left(r_{1}+r_{2}\right)\\\beta &=\textstyle {\frac {1}{2}}\left(r_{1}-r_{2}\right).\end{aligned}}$

Thus, solving for the resolvents gives the original roots.

Formally, the resolvents are called the discrete Fourier transform (DFT) of order 2, and the transform can be expressed by the matrix $\left({\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}}\right),$ with inverse matrix $\left({\begin{smallmatrix}1/2&1/2\\1/2&-1/2\end{smallmatrix}}\right).$ The transform matrix is also called the DFT matrix or Vandermonde matrix.

Now $r_{1}=\alpha +\beta$ is a symmetric function in $\alpha$ and $\beta ,$ so it can be expressed in terms of p and q, and in fact $r_{1}=-p,$ as noted above. But $r_{2}=\alpha -\beta$ is not symmetric, since switching $\alpha$ and $\beta$ yields $-r_{2}=\beta -\alpha$ (formally, this is termed a group action of the symmetric group of the roots). Since $r_{2}$ is not symmetric, it cannot be expressed in terms of the polynomials p and q, as these are symmetric in the roots and thus so is any polynomial expression involving them. Changing the order of the roots only changes $r_{2}$ by a factor of $-1,$ and thus the square $\scriptstyle r_{2}^{2}=(\alpha -\beta )^{2}$ is symmetric in the roots, and thus expressible in terms of p and q. Using the equation

$(\alpha -\beta )^{2}=(\alpha +\beta )^{2}-4\alpha \beta \!$

yields

$r_{2}^{2}=p^{2}-4q\!$

and thus

$r_{2}=\pm {\sqrt {p^{2}-4q}}.\!$

If one takes the positive root, breaking symmetry, one obtains:

${\begin{aligned}r_{1}&=-p\\r_{2}&={\sqrt {p^{2}-4q}}\end{aligned}}$

and thus

${\begin{aligned}\alpha &=\textstyle {\frac {1}{2}}\left(-p+{\sqrt {p^{2}-4q}}\right)\\\beta &=\textstyle {\frac {1}{2}}\left(-p-{\sqrt {p^{2}-4q}}\right)\end{aligned}}$

Thus the roots are

$\textstyle {\frac {1}{2}}\left(-p\pm {\sqrt {p^{2}-4q}}\right)$

which is the quadratic formula. Substituting $\scriptstyle p={\tfrac {b}{a}},q={\tfrac {c}{a}}\!$ yields the usual form for when a quadratic is not monic. The resolvents can be recognized as $\scriptstyle {\frac {r_{1}}{2}}={\frac {-p}{2}}={\frac {-b}{2a}}\!$ being the vertex, and $\scriptstyle r_{2}^{2}=p^{2}-4q\!$ is the discriminant (of a monic polynomial).

A similar but more complicated method works for cubic equations, where one has three resolvents and a quadratic equation (the "resolving polynomial") relating $r_{2}$ and $r_{3},$ which one can solve by the quadratic equation, and similarly for a quartic (degree 4) equation, whose resolving polynomial is a cubic, which can in turn be solved. The same method for a quintic equation yields a polynomial of degree 24, which does not simplify the problem, and in fact solutions to quintic equations in general cannot be expressed using only roots.

See also

• Discriminant
• Fundamental theorem of algebra

External links

• Quadratic formula calculator
• Quadratic formula calculator Online
• Alternative formula (Wolfram)