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Algebra 3
Multiplying expressions, solving a quadratic by factoring, Complex Numbers, quadratic equation, using the quadratic ,Equation to solve 2nd degree polynomials, Completing the square, Quadratic Formula (proof) , Quadratic Formula Quadratic Inequalities, Introduction to functions, Functions, Domain of a function, Proof: log a + log b = log ab, Proof: A (log B) = log (B^A), log A - log B = log (A/B), Proofs of the logarithm properties: A (log B) = log (B^A) and log A - log B = log (A/B), Proof: log_a (B) = (log_x (B))/ (log_x (A)), Algebraic Long Division, Conic Sections, Circles, Ellipses, Hyperbolas, Identifying Conics
Algebra 4

Foci of an Ellipse, Foci of a Hyperbola, Proof: Hyperbola Foci, Partial Fraction Expansion, Parabola Focus and Directrix, Two Passing Bicycles Word Problem, Passing Trains problem, Overtaking Word Problem


Quadratic equation

In mathematics, a quadratic equation is a polynomial equation of the second degree. The general form is where x represents a variable, and a, b, and c, constants, with a ≠ 0. (If a = 0, the equation becomes a linear equation.)

The constants a, b, and c, are called respectively, the quadratic coefficient, the linear coefficient and the constant term or free term. The term "quadratic" comes from quadratus, which is the Latin word for "square." Quadratic equations can be solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula.  One common use of quadratic equations is computing trajectories in projectile motion.
 
 

Quadratic formula

A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real. The roots are given by the quadratic formula:
x=frac{-b pm sqrt {b^2-4ac}}{2a}

where the symbol "±" indicates that both

frac{-b + sqrt {b^2-4ac}}{2a} and frac{-b - sqrt {b^2-4ac}}{2a}

are solutions for x.

Discriminant

Example discriminant signs
<0: x2+12
=0: −43x2+43x13
>0: 32x2+12x43

In the above formula, the expression underneath the square root sign is called the discriminant of the quadratic equation, and is often represented using an upper case Greek Delta:

Delta = b^2 - 4ac.,

A quadratic equation with real coefficients can have either one or two distinct real roots, or two distinct complex roots. In this case the discriminant determines the number and nature of the roots. There are three cases:

  • If the discriminant is positive, there are two distinct roots, both of which are real numbers.
For quadratic equations with integer coefficients, if the discriminant is a perfect square, then the roots are rational numbers—in other cases they may be quadratic irrationals.
  • If the discriminant is zero, there is exactly one distinct real root, sometimes called a double root:
 x = -frac{b}{2a} . ,!
  • If the discriminant is negative, there are no real roots. Rather, there are two distinct (non-real) complex roots, which are complex conjugates of each other:
frac{-b}{2a} + i frac{sqrt {4ac - b^2}}{2a} and  frac{-b}{2a} - i frac{sqrt {4ac - b^2}}{2a},
where i is the imaginary unit.

Thus the roots are distinct if and only if the discriminant is non-zero, and the roots are real if and only if the discriminant is non-negative.

 

Geometry

For the quadratic function:
f (x) = x2x − 2 = (x + 1)(x − 2) of a real variable x, the x-coordinates of the points where the graph intersects the x-axis, x = −1 and x = 2, are the roots of the quadratic equation: x2x − 2 = 0.

The solutions of the quadratic equation

ax^2+bx+c=0,,

are also the zeros of the quadratic function:

f(x) = ax^2+bx+c,,

since they are the values of x for which

f(x) = 0.,

If a, b, and c are real numbers and the domain of f is the set of real numbers, then the zeros of f are exactly the x-coordinates of the points where the graph touches the x-axis.

It follows from the above that, if the discriminant is positive, the graph touches the x-axis at two points, if zero, the graph touches at one point, and if negative, the graph does not touch the x-axis.

Quadratic factorization

The term

x - r,

is a factor of the polynomial

ax^2+bx+c,

if and only if r is a root of the quadratic equation

ax^2+bx+c=0.
It follows from the quadratic formula that
 
ax^2+bx+c = a left( x - frac{-b + sqrt {b^2-4ac}}{2a} right) left( x - frac{-b - sqrt {b^2-4ac}}{2a} right).

In the special case (b2 = 4ac) where the quadratic has only one distinct root (i.e. the discriminant is zero), the quadratic polynomial can be factored as

ax^2+bx+c = a left( x + frac{b}{2a} right)^2.,!


Application to higher-degree equations

Certain higher-degree equations can be brought into quadratic form and solved that way. For example, the 6th-degree equation in x:

x^6 - 4x^3 + 8 = 0,

can be rewritten as:

(x^3)^2 - 4(x^3) + 8 = 0,,

or, equivalently, as a quadratic equation in a new variable u:

u^2 - 4u + 8 = 0,,

where

u = x^3.,

Solving the quadratic equation for u results in the two solutions:

u = 2 pm 2i,.

Thus

x^3 = 2 pm 2i,.

Concentrating on finding the three cube roots of 2 + 2i – the other three solutions for x will be their complex conjugates – rewriting the right-hand side using Euler's formula:

x^3 = 2^{tfrac{3}{2}}e^{tfrac{1}{4}pi i} = 2^{tfrac{3}{2}}e^{tfrac{8k+1}{4}pi i},

(since e2kπi = 1), gives the three solutions:

x = 2^{tfrac{1}{2}}e^{tfrac{8k+1}{12}pi i},,~k = 0, 1, 2,.

Using Eulers' formula again together with trigonometric identities such as cos(π/12) = (√2 + √6) / 4, and adding the complex conjugates, gives the complete collection of solutions as:

x_{1,2} = -1 pm i,,
x_{3,4} = frac{1 + sqrt{3}}{2} pm frac{1 - sqrt{3}}{2}i,

and

x_{5,6} = frac{1 - sqrt{3}}{2} pm frac{1 + sqrt{3}}{2}i.,

Derivation 

 


The quadratic formula can be derived by the method of completing the square, so as to make use of the algebraic identity:

x^2+2xh+h^2 = (x+h)^2.,!

Dividing the quadratic equation

ax^2+bx+c=0 ,!

by a (which is allowed because a is non-zero), gives:

x^2 + frac{b}{a} x + frac{c}{a}=0,,!

or

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

The quadratic equation is now in a form to which the method of completing the square can be applied. To "complete the square" is to add a constant to both sides of the equation such that the left hand side becomes a complete square:

x^2+frac{b}{a}x+left( frac{1}{2}frac{b}{a} right)^2 =-frac{c}{a}+left( frac{1}{2}frac{b}{a} right)^2,!

which produces

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

The right side can be written as a single fraction, with common denominator 4a2. This gives

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

Taking the square root of both sides yields

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

Isolating x, gives

x=-frac{b}{2a}pmfrac{sqrt{b^2-4ac }}{2a}=frac{-bpmsqrt{b^2-4ac }}{2a}.

By Lagrange resolvents

An alternative way of deriving the quadratic formula is via the method of Lagrange resolvents, which is an early part of Galois theory. A benefit of this method is that it generalizes to give the solution of cubic polynomials and quartic polynomials, and leads to Galois theory, which allows one to understand the solution of polynomials 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 α and β and the values of p and q will not change: one says that p and q are symmetric polynomials in α and β. In fact, they are the elementary symmetric polynomials – any symmetric polynomial in α and β can be expressed in terms of α + β and αβ. 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 Sn. 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 α and β, consider their sum and difference:

begin{align}
r_1 &= alpha + beta
r_2 &= alpha - beta.
end{align}

These are called the Lagrange resolvents of the polynomial; notice that these depend 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{align}
alpha &= textstyle{frac{1}{2}}left(r_1+r_2right)
beta &= textstyle{frac{1}{2}}left(r_1-r_2right).
end{align}

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 & -1end{smallmatrix}right), with inverse matrix left(begin{smallmatrix}1/2 & 1/2 1/2 & -1/2end{smallmatrix}right). The transform matrix is also called the DFT matrix or Vandermonde matrix.

Now r1 = α + β is a symmetric function in α and β, so it can be expressed in terms of p and q, and in fact r1 = − p, as noted above. Contrariwise, r2 = α − β is not symmetric, since switching α and β yields r2 = β − α (formally, this is termed a group action of the symmetric group of the roots). Since r2 is not symmetric, it cannot be expressed in terms of the roots p and q, as these are symmetric in the roots and thus so is any polynomial expression involving them. However, changing the order of the roots only changes r2 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 - 4alphabeta!

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{align}
r_1 &= -p
r_2 &= sqrt{p^2 - 4q}
end{align}

and thus

begin{align}
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{align}

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 r2 and r3, 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. However, 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.

 

Alternative formula

In some situations it is preferable to express the roots in an alternate form.

x =frac{2c}{-b mp sqrt {b^2-4ac }} .

This alternative requires c to be nonzero; for, if c is zero, the formula correctly gives zero as one root, but fails to give any second, non-zero root. Instead, one of the two choices for ∓ produces a division by zero, which is undefined.

The roots are the same regardless of which expression we use; the alternate form is merely an algebraic variation of the common form:
 
begin{align}
frac{-b + sqrt {b^2-4ac }}{2a}
 &{}= frac{-b + sqrt {b^2-4ac }}{2a} cdot frac{-b - sqrt {b^2-4ac }}{-b - sqrt {b^2-4ac }}
 &{}= frac{4ac}{2a left ( -b - sqrt {b^2-4ac} right ) }
 &{}=frac{2c}{-b - sqrt {b^2-4ac }}.
end{align}

The alternative formula can reduce loss of precision in the numerical evaluation of the roots, which may be a problem if one of the roots is much smaller than the other in absolute magnitude. The problem of c possibly being zero can be avoided by using a mixed approach:

x_1 = frac{-b - sgn (b) ,sqrt {b^2-4ac}}{2a},
x_2 = frac{c}{ax_1}.

Here sgn denotes the sign function.

Monic form

Dividing the quadratic equation by a gives the simplified monic form of

x2 + px + q = 0

where p = ba and q = ca. This in turn simplifies the root and discriminant equations somewhat to

x = frac{1}{2} left( -p pm sqrt{p^2 - 4q} ,right)
Δ = p2 − 4q

(Cf. the Lagrange resolvents derivation method above.)

 

Floating point implementation

A careful floating point computer implementation differs a little from both forms to produce a robust result. Assuming the discriminant, b2 − 4ac, is positive and b is nonzero, the code will be something like the following.

t := -tfrac12 big( b + sgn(b) sqrt{b^2-4ac} big) ,!
x_1 := t/a ,!
x_2 := c/t ,!

Here sgn(b) is the sign function, where sgn(b) is 1 if b is positive and −1 if b is negative; its use ensures that the quantities added are of the same sign, avoiding catastrophic cancellation. The computation of x2 uses the fact that the product of the roots is c/a.

See Numerical Recipes in C, Section 5.6: "Quadratic and Cubic Equations".

Viète's formulas

Viète's formulas give a simple relation between the roots of a polynomial and its coefficients. In the case of the quadratic polynomial, they take the following form:

 x_1 + x_2 = -frac{b}{a}

and

 x_1 cdot x_2 = frac{c}{a}.

The first formula above yields a convenient expression when graphing a quadratic function. Since the graph is symmetric with respect to a vertical line through the vertex, when there are two real roots the vertex’s x-coordinate is located at the average of the roots (or intercepts). Thus the x-coordinate of the vertex is given by the expression:

 x_V = frac {x_1 + x_2} {2} = -frac{b}{2a}.

The y-coordinate can be obtained by substituting the above result into the given quadratic equation, giving

 y_V = - frac{b^2}{4a} + c = - frac{ b^2 - 4ac} {4a}.

Generalizations

The formula and its derivation remain correct if the coefficients a, b and c are complex numbers, or more generally members of any field whose characteristic is not 2. (In a field of characteristic 2, the element 2a is zero and it is impossible to divide by it.)

The symbol

pm sqrt {b^2-4ac}

in the formula should be understood as "either of the two elements whose square is b2 − 4ac, if such elements exist". In some fields, some elements have no square roots and some have two; only zero has just one square root, except in fields of characteristic 2. Note that even if a field does not contain a square root of some number, there is always a quadratic extension field which does, so the quadratic formula will always make sense as a formula in that extension field.

Characteristic 2

In a field of characteristic 2, the quadratic formula, which relies on 2 being a unit, does not hold. Consider the monic quadratic polynomial

displaystyle x^{2} + bx + c

over a field of characteristic 2. If b = 0, then the solution reduces to extracting a square root, so the solution is

displaystyle x = sqrt{c}

and note that there is only one root since

displaystyle -sqrt{c} = -sqrt{c} + 2sqrt{c} = sqrt{c}.

In summary,

displaystyle x^{2} + c = (x + sqrt{c})^{2}.

See quadratic residue for more information about extracting square roots in finite fields.

In the case that b ≠ 0, there are two distinct roots, but if the polynomial is irreducible, they cannot be expressed in terms of square roots of numbers in the coefficient field. Instead, define the 2-root R(c) of c to be a root of the polynomial x2 + x + c, an element of the splitting field of that polynomial. One verifies that R(c) + 1 is also a root. In terms of the 2-root operation, the two roots of the (non-monic) quadratic ax2 + bx + c are

frac{b}{a}Rleft(frac{ac}{b^2}right)

and

frac{b}{a}left(Rleft(frac{ac}{b^2}right)+1right).

For example, let a denote a multiplicative generator of the group of units of F4, the Galois field of order four (thus a and a + 1 are roots of x2 + x + 1 over F4). Because (a + 1)2 = a, a + 1 is the unique solution of the quadratic equation x2 + a = 0. On the other hand, the polynomial x + ax + 1 is irreducible over F4, but splits over F16, where it has the two roots ab and ab + a, where b is a root of x2 + x + a in F16.

This is a special case of Artin-Schreier theory.

 

 
 
 
 
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