most equadratic equations have ?

Question

one solution?
two solutions?
four solutions?
an unlimited solution?

Answer 1

two solutions.

When asked to find the solution to a quadratic equation, what you're really doing finding where the graph of that equation (a parabola) intersects the x axis. If the vertex of the parabola is on the x axis, it will have one solution. If the vertex is above the x axis and opens upward, it will have no solutions, just as if the vertex were below the x axis and it opens downwards. But it will have two solutions if the vertex is below the x axis and opens upward, or if the vertex is above the x axis and opens downward.

Answer 2

Most quadratics will have 0, 1 or 2 solutions

Okay - everybody wants to be picky.

All quadratics will have 2 solutions, but in some cases the solutions will be imaginary.

Regarding real solutions on a cartesian grid a quadratic function will have:

No solutions if the parabola's vertex doesn't reach the x-axis

One solution if the x axis is tangential to the parabola's vertex

and Two solutions if the vertex of the parabola lies beyond the x-axis compared with the rest of the parabola.

This can be established by the discriminant of the quadratic function

Discriminant = b^2 - 4ac
if the function is of the form f(x) = ax^2 + bx + c

If the discriminant is 0 then there is 1 root (or a double root, both values being the same)
If the discriminant is negative there are no roots (or two imaginary roots)
If the discriminant is positive then there are 2 real roots given by

x = (-b +/- SQRT(b^2-4ac))/2a

Answer 3

you have to define what consitutes a solution. In the real numbers some quadratics have zero solutions, but not so in the complex plane.
Since you said most it is interesting to assume the real numbers.
There are an infinate number of all 3, however it does NOT follwo that given any arbitrary quadtratic all 3 are equally likely. It would be relatively uncommon, for example, for a quadratic to have exactly one solution. This would require the parabola simply just touching the line at a single point, so for uncountably infinate numbers A and B there would only exist one number C that gives one solution.
There should be equal number of zero and 2 solutions for the same reason that there is a negative number associated with every positive number, if an equation has zero solutions simply mulitply it by -1 and you get two solutions. So i would saty 50/50 chance of 0 or 2 solutions, and statistically zero chance of one solution, althought there are an infinate number of quadratics with one solution.

Answer 4

Most have 2. Some have zero and some have 1. None can have more than 2

x^2 = -6 has zero real solutions
x^2 -4x +4 = 0 has one real solution (x=2)
x^2 + 7x + 12 = 0 has two real solutions (x= -4 and x = -3)

Most quadratic problems will have two solutions

Answer 5

strange question because the word "most" should be ALL. at any rate, ALL quadratic equations have exactly 2 solutions.


why is everyone talking about the graph? the "EQUATION" always has exactly 2 solutions. The fundamental theorem of algebra demands it.

Furthermore, for the graph in the Real Field if it just touches the x-axis (this is called OSCULATING) there STILL EXISTS EXACTLY 2 solutions. The root has multiplicity 2.

Answer 6

TWO solutions... know why?
Solutions are of the highest degree of it uknown.

Answer 7

MOST QUADRATIC EQUATIONS WILL HAVE TWO SOLUTIONS BUT SOMTIMES IN VERY RARE CASES LIKE IF YOU GET IMAGINARY NUMBERS THEY HAVE NO SOLUTIONS AND SOMETIMES THEY WIL HAVE ONE SOLUTION.

Answer 8

Quadratic Equationa

2 solutions

- - - - - - - - - -s-

Answer 9

Quadratic formula

A quadratic equation with real (or complex) coefficients has two (but not necessarily distinct) solutions, called roots, which may be real or complex, given by the quadratic formula:

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

where the symbol "±" indicates that both

x_+ = \frac{-b + \sqrt {b^2-4ac}}{2a} and \ x_- = \frac{-b - \sqrt {b^2-4ac}}{2a},

are solutions.

Discriminant

In the above formula, the term underneath the square root sign:

b^2 - 4ac,\,\!

is called the discriminant of the quadratic equation.

A quadratic equation with real coefficients can have either one or two distinct roots, each of which is either real or complex. 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 root, and that root is a real number. Sometimes called a double root, its value is:

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:

x_+ = \frac{-b}{2a} + i \left ( \frac{\sqrt {4ac - b^2}}{2a} \right ) and x_- = \frac{-b}{2a} - i \left ( \frac{\sqrt {4ac - b^2}}{2a} \right ),

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) = x2 − x − 2 = (x + 1)(x − 2) of a real variable x, the x-coordinates of the points where the graph touches the x-axis, x = −1 and x = 2, are the roots of the quadratic equation: x2 − x − 2 = 0.
For the quadratic function:
f (x) = x2 − x − 2 = (x + 1)(x − 2) of a real variable x, the x-coordinates of the points where the graph touches the x-axis, x = −1 and x = 2, are the roots of the quadratic equation: x2 − x − 2 = 0.

The roots 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.

[edit] 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 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.\,\!

[edit] Application to higher-degree equations

Certain higher-degree equations may be quadratic in form, such as:

2x^6 + 3x^3 + 5 = 0,\,

which can be written as

2u^2 + 3u + 5 = 0, \

where

u = x^3 \.

Note that the highest exponent is twice the value of the exponent of the middle term. This equation may be resolved directly or with a simple substitution, using the methods that are available for the quadratic, such as factoring, the quadratic formula, or completing the square.

Generally speaking, if the polynomial is quadratic in some variable u where

u = x^n \,\;

then the quadratic equation can be used to help find solutions.

History

On clay tablets dated between 1800 BC and 1600 BC, the ancient Babylonians left the earliest evidence of the discovery of quadratic equations, and also gave early methods for solving them. Indian mathematician Baudhayana who wrote a Sulba Sutra in ancient India circa 8th century BC first used quadratic equations of the form ax2 = c and ax2 + bx = c and also gave methods for solving them.

Babylonian mathematicians from circa 400 BC and Chinese mathematicians from circa 200 BC used the method of completing the square to solve quadratic equations with positive roots, but did not have a general formula. Euclid produced a more abstract geometrical method around 300 BC. The Bakshali Manuscript written in India between 200 BC and 400 AD introduced the general algebraic formula for solving quadratic equations, and also introduced quadratic indeterminate equations (origin of type ax/c = y).

The first mathematician to have found negative solutions with the general algebraic formula was Brahmagupta (India, 7th century). Muḥammad ibn Mūsā al-Ḵwārizmī (Persia, 9th century) developed a set of formulae that worked for positive solutions. Abraham bar Hiyya Ha-Nasi (also known by the Latin name Savasorda) introduced the complete solution to Europe in his book Liber embadorum in the 12th century. Bhaskara II (1114-1185), an Indian mathematician-astronomer, also known as Bhaskara II and Bhaskara Achārya ("Bhaskara the teacher") solved quadratic equations with more than one unknown and is considered the originator of the equation. [1]

Shridhara (India, 9th century) was one of the first mathematicians to give a general rule for solving a quadratic equation. His original work is lost but Bhaskara II later quotes Shridhara's rule:

Multiply both sides of the equation by a known quantity equal to four times the coefficient of the square of the unknown; add to both sides a known quantity equal to the square of the coefficient of the unknown; then take the square root. [2]

Derivation

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

x^2+2xy+y^2 = (x+y)^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} \qquad (1)

The quadratic equation is now in a form in which the method of completing the square can be applied. To "complete the square" is to find some constant k such that

x^2 + \frac{b}{a}x + k = x^2+2xy+y^2,\,\!

for another constant y. In order for these equations to be true,

y = \frac{b}{2a}\,\!

and

k = y^2,\,\!

thus

k = \frac{b^2}{4a^2}.\,\!

Adding this constant to equation (1) produces

x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}.\,\!

The left side is now a perfect square because

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

\left|x+\frac{b}{2a}\right| = \frac{\sqrt{b^2-4ac\ }}{|2a|}\Rightarrow x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac\ }}{2a}.

Isolating x, gives

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

Alternative formula

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

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

However, it imposes the additional requirement that c be nonzero. If c is zero, this formula correctly gives zero as one root, but fails to give any second, non-zero root. (When c is zero we have division of zero by zero, which is indeterminate.)

The actual values of the roots must be the same regardless of which expression we use, so the alternate form is merely an algebraic variation of the common form. For example,

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

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 := -(b + \sgn(b) \sqrt{b^2-4ac})/2 \,\!
r_{1} := t/a \,\!
r_{2} := c/t \,\!

Here sgn(b) is the sign function, giving +1 if b is positive and −1 if b is negative; its use ensures that we always add two quantities of the same sign, avoiding catastrophic cancellation. The computation of r2 uses the fact that the product of the roots is c/a.

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_+ + x_- = -\frac{b}{a}

and

x_+ \cdot x_- = \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_+ + x_-} {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}.

[edit] 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 x2 + bx + c over a field of characteristic 2. If b = 0, then the solution reduces to extracting a square root, so the solution is x = √c and note that there is only one root since –√c = –√c + 2√c = √c. In summary, x2 + c = (x + √c)2. Confer 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}R\left(\frac{ac}{b^2}\right)

and

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

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.