Quadratic equation/Advanced

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In mathematics, or more specifically algebra, a quadratic equation is one involving only polynomials of the second degree. Quadratic equations are a common part of mathematical solutions to real-world problems in a huge variety of situations. Quadratic equations occurring in applications typically involve real number coefficients. However, one can algebraically manipulate polynomial equations in the usual way as long as the coefficients can be added and multiplied together. Please see the main page for a discussion of polynomials with real coefficients.

The most general mathematical context that deals with systems of objects that can be added and multiplied together is ring theory. One can define polynomials, and in particular quadratic polynomial equations, as long as the coefficients are in a ring. The real numbers is an example of a ring. Another example, important in coding theory, is polynomials with coefficients in the ring $\mathbb{Z}_2 = \{ \, \overline{0}, \overline{1} \, \}$ . You add and multiply in this ring in the same way you add or multiply the integers $\{ \, 0, 1, \, \}$ with one exception: since $\mathbb{Z}_2$ does not have a "two" in it, we set $\overline{1} + \overline{1} = \overline{0}$ .

[edit] Solutions of quadratic equations

When working with polynomials over a specific ring $R$ , one usually looks for solutions in the same ring $R$ . The main exception to this is the most common case, where a polynomial has integer coefficients but one desires real number solutions. If, instead, one demands solutions of the same type as the polynomial coefficients, namely integers, the equation becomes a Diophantine equation. In this article, we assume that the desired solutions are in the same ring that the coefficients are drawn from.

Every polynomial equation with coefficients in a ring $R$ can be put into the form:

$ax^2+bx+c=0\,$

with a, b and c in $R$ and $a\not=0$ . When the coefficients are real numbers, the quadratic formula specifies the roots of this equation as

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

If $R$ is an arbitrary ring, however, there are several problems with this formula. The derivation of the formula typically involves completing the square. Looking at the derivation on the main page, the first problem arises when the factoring is performed. If $R$ is not commutative, this step is not valid. For the rest of this section, we assume that $R$ is a commutative ring.

The next problem in the derivation of the quadratic formula for commutative rings arises when the square root is taken. This is a problem already familiar from the situation where the coefficients are real numbers: if the discriminant is negative, then there are no real solutions. Otherwise, there are one or two solutions. If the coefficients are in a general commutative ring, the discriminant can have 0, 1, 2, or even more square roots.

For example, let $R$ be the ring of integers modulo 15,

$\mathbb{Z}_{15} = \{ \, \overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4}, \overline{5}, \overline{6}, \overline{7}, \overline{8}, \overline{9}, \overline{10}, \overline{11}, \overline{12}, \overline{13}, \overline{14} \, \}$

Multiplication in this ring is performed by multiplying the representative integers and taking the remainder upon dividing the result by 15. Using this procedure, we calculate the squares of the elements in this ring

$\overline{0}^2 = \overline{0} \qquad \qquad \, \, \, \overline{5}^2 = \overline{25} = \overline{10} \qquad \qquad \overline{10}^2 = \overline{100} = \overline{10}$

$\overline{1}^2 = \overline{1} \qquad \qquad \, \, \, \overline{6}^2 = \overline{36} = \overline{6} \qquad \qquad \, \, \, \overline{11}^2 = \overline{121} = \overline{1}$

$\overline{2}^2 = \overline{4} \qquad \qquad \, \, \, \overline{7}^2 = \overline{49} = \overline{4} \qquad \qquad \, \, \, \overline{12}^2 = \overline{144} = \overline{9}$

$\overline{3}^2 = \overline{9} \qquad \qquad \, \, \, \overline{8}^2 = \overline{64} = \overline{4} \qquad \qquad \, \, \, \overline{13}^2 = \overline{169} = \overline{4}$

$\overline{4}^2 = \overline{16} = \overline{1} \qquad \overline{9}^2 = \overline{81} = \overline{6} \qquad \qquad \, \, \, \overline{14}^2 = \overline{196} = \overline{1}$

We see that the elements $\overline{2}, \overline{3}, \overline{5}, \overline{7}, \overline{8}, \overline{11}, \overline{12}, \overline{13}$ and $\overline{14}$ have no square roots in the ring, while $\overline{1}$ has four distinct square roots!

When working with real coefficients, we can work around the case where the discriminant is negative by allowing solutions in the larger ring of complex numbers. In the above example, it is not clear how to enlarge the ring of integers modulo 15 to ensure that the discriminant has a square root. However, when the ring of coefficients is a field, one can ensure that there are solutions for every quadratic equation by allowing solutions to be in an algebraic closure of the field (see below).

When $R$ is an arbitrary commutative ring and we seek solutions in the same ring, we simply say that the equation is unsolvable over $R$ if the discriminant does not have a square root in the ring.

The final problem with the derivation of the quadratic formula over a general commutative ring is the final division by $2 a$ . Since division is not a ring operation, we must be careful to view this step as a multiplication by the multiplicative inverse of $2 a$ . It is very possible that $2 a$ is not invertible. In particular, this happens if the characteristic of the ring is divisible by two . However, if both 2 and a are both units in $R$ , then this step can be performed.

In summary, the quadratic formula is valid for quadratic equations over a general ring $R$ with the following caveats:

$R$ must be commutative
$2$ and $a$ must be units in $R$ , so in particular $R$ must not have characteristic divisible by two.
The term $\sqrt{b^2-4ac}$ must be allowed to take all possible square roots (sometimes more than two!) in $R$

[edit] Example

Let us use the quadratic formula to find all solutions of the quadratic equation

$\overline{4} x^2 + \overline{2} x + \overline{3} = \overline{0}$

over the ring of integers modulo 15. As discussed above, we cannot apply the formula unless 2a is a unit in the ring. In this example, $2a=\overline{8}$ , which is a unit in $\mathbb{Z}_{15}$ with inverse $\overline{2}$ (since $\overline{8} \cdot \overline{2} = \overline{16} = \overline{1}$ ). The quadratic formula then says

$x = \left( -\overline{2} + \sqrt{\overline{2}^2 - 4 \cdot\overline{4} \cdot \overline{3}} \right) * (2 \cdot \overline{4})^{-1} = \left( - \overline{2} + \sqrt{\overline{1}} \right) * \overline{2}$

Recalling that we must consider all possible square roots here, and consulting the table of squares in $\mathbb{Z}_{15}$ above, we see that there are four possible values for $\sqrt{\overline{1}}$ : $\overline{1}, \overline{4}, \overline{11}$ and $\overline{14}$ . Substituting each of these in for the square root above and simplifying, we find that the four solutions $x = \overline{3}, \overline{4}, \overline{9}$ , and $\overline{13}$ . Substituting these back into the original equation shows that they are indeed solutions.

[edit] General fields

Let Δ be the discriminant,

$\Delta = b^2 - 4ac .\,$

This pair of solutions, which may be verified by completing the square, are valid when the characteristic of F is not 2: we shall assume this for now and deal with binary fields below.

If Δ is a square in F then the quadratic equation splits completely in F: that is, both roots lie in F.

In Δ is not a square in F then the field extension $F(\sqrt\Delta)$ is quadratic over F: both roots of the equation lie in the extension, which is thus a splitting field for the equation and hence a Galois extension.

We observe that in this case, the quadratic equations is soluble by radicals: in this case, square roots.

[edit] Characteristic two

In the case of binary fields, extensions by square roots are not the most general form of quadratic extension. The map $X \mapsto X^2$ is always injective, and in the case of finite fields it is therefore also surjective (it is the Frobenius automorphism).

To obtain the most general quadratic extension, consider the Artin-Schreier polynomial

$A_\alpha(X) = X^2 + X + \alpha \,$

for α in F. The function $A : X \mapsto X^2 + X$ is two-to-one since $A (x) = A (x + 1)$ . It is in fact $\mathbf{F}_2$ -linear on F as a vector space.

[edit] Finite fields

Suppose that F is finite, of characteristic two. The Frobenius map is an automorphism and so its inverse, the square root map is defined everywhere, and square roots do not generate any non-trivial extensions.

If F is finite, then A is exactly 2-to-1 and the image of A is a $\mathbf{F}_2$ -subspace of codimension 1. There is always some element α of F not in the image of A, and so the corresponding Artin-Schreier polynomial has no root in F: it is therefore an irreducible polynomial and the quotient ring $F[X]/\langle A_\alpha(X) \rangle$ is a field which is a quadratic extension of F. Since finite fields of the same order are unique up to isomorphism, we may say that this is "the" quadratic extension of F. As before, both roots of the equation lie in the extension, which is thus a splitting field for the equation and hence a Galois extension: in this case the roots are of the form $\beta,~\beta+1$ .

Quadratic equation/Advanced

Contents

[edit] Solutions of quadratic equations

[edit] Example

[edit] General fields

[edit] Characteristic two

[edit] Finite fields

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