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Not for all types of equations. But always in second degree equations they do.

Consider a third degree equation with 3 different roots. Obviously, one of the roots can not be in a pair.

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โˆ™ 2012-08-11 09:01:40
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Algebra

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A polynomial of degree zero is a constant term

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Q: Complex roots occur in conjugate pairs?
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Related questions

What is the relationship between the degree of a polynomial and the number of roots it has?

In answering this question it is important that the roots are counted along with their multiplicity. Thus a double root is counted as two roots, and so on. The degree of a polynomial is exactly the same as the number of roots that it has in the complex field. If the polynomial has real coefficients, then a polynomial with an odd degree has an odd number of roots up to the degree, while a polynomial of even degree has an even number of roots up to the degree. The difference between the degree and the number of roots is the number of complex roots which come as complex conjugate pairs.


Can you have a quadratic function with one real root and one complex root?

No you can not. Complex roots appear as conjugates. if a root is complex so is its conjugate. so either the roots are real or are both coplex.


The graph of a polynomial changes direction twice and has only one root What can you say about the polynomial?

It is a polynomial of odd power - probably a cubic. It has only one real root and its other two roots are complex conjugates. It could be a polynomial of order 5, with two points of inflexion, or two pairs of complex conjugate roots. Or of order 7, etc.


What is the relationship between the degree and the number of roots counting multiplicities?

In the complex field, the two numbers are the same. If you restrict yourself to real solutions, the relationship is as follows: A polynomial of degree p has p-2k real solutions where k is an integer such that p-2k is non-negative. [There will be 2k pairs of complex conjugate roots.]


How many roots in a radical problem if the index is odd?

An odd number. In the complex field, the number of roots is the same as the index. Complex (non-real) roots come in pairs (complex conjugates) so the number of real roots will also be odd.


Do every polynomial function has at least one complex zero?

No. Complex zeros always come in conjugate pairs. So if a+bi is one zero, then a-bi is also a zero.The fundamental theorem of algebra says"Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers."If you want to know how many complex root a given polynomial has, you might consider finding out how many real roots it has. This can be done with Descartes Rules of signsThe maximum number of positive real roots can be found by counting the number of sign changes in f(x). The actual number of positive real roots may be the maximum, or the maximum decreased by a multiple of two.The maximum number of negative real roots can be found by counting the number of sign changes in f(-x). The actual number of negative real roots may be the maximum, or the maximum decreased by a multiple of two.Complex roots always come in pairs. That's why the number of positive or number of negative roots must decrease by two. Using the two rules for positive and negative signs along with the fact that complex roots come in pairs, you can determine the number of complex roots.


Which polynomial has rational coefficients a leading leading coefficient of 1 and the zeros at 2-3i and 4?

There cannot be such a polynomial. If a polynomial has rational coefficients, then any complex roots must come in conjugate pairs. In this case the conjugate for 2-3i is not a root. Consequently, either (a) the function is not a polynomial, or (b) it does not have rational coefficients, or (c) 2 - 3i is not a root (nor any other complex number), or (d) there are other roots that have not been mentioned. In the last case, the polynomial could have any number of additional (unlisted) roots and is therefore indeterminate.


Can the roots of a quadratic be listed as ordered pairs?

In the case of real roots, you could, but the second part of the ordered pair (the ordinate) will always be zero, so there is not much point.In the case of complex roots (or real roots in the complex field), you could list them as ordered pairs: with (a, b) representing a + bi where i is the imaginary square root of -1..


If the roots are real then which type of vibrations will occur in damped systems?

real roots= Overdamped equal roots= critically damped complex roots /imaginary roots = Underdamped


Is it true that the degree of polynomial function determine the number of real roots?

Sort of... but not entirely. Assuming the polynomial's coefficients are real, the polynomial either has as many real roots as its degree, or an even number less. Thus, a polynomial of degree 4 can have 4, 2, or 0 real roots; while a polynomial of degree 5 has either 5, 3, or 1 real roots. So, polynomial of odd degree (with real coefficients) will always have at least one real root. For a polynomial of even degree, this is not guaranteed. (In case you are interested about the reason for the rule stated above: this is related to the fact that any complex roots in such a polynomial occur in conjugate pairs; for example: if 5 + 2i is a root, then 5 - 2i is also a root.)


Which arithmetic operation requires the use of the complex conjugate?

It can be used as a convenient shortcut to calculate the absolute value of the square of a complex number. Just multiply the number by its complex conjugate.I believe it has other uses as well.


What is the cube root of -2?

It's the same as the negative of the cube root of 2.


How many complex roots does 6x4 plus x3 - 5 equals 0 have?

It has two complex roots.


How could you use Descartes' rule to predict the number of complex roots to a polynomial?

Descartes' rule of signs will not necessarily tell exact number of complex roots, but will give an idea. The Wikipedia article explains it pretty well, but here is a brief explanation:It is for single variable polynomials.Order the polynomial in descending powers [example: f(x) = ax³ + bx² + cx + d]Count number of sign changes between consecutive coefficients. This is the maximum possible real positive roots.Let function g(x) = f(-x), count number of sign changes, which is maximum number of real negative roots.Note these are maximums, not the actual numbers. Let p = positive maximum and q = negative maximum. Let m be the order (maximum power of the variable), which is also the total number of roots.So m - p - q = minimum number of complex roots. Note complex roots always occur in pairs, so number of complex roots will be {0, 2, 4, etc}.


What are complex root?

The complex roots of an equation are the complex numbers that are solutions to the equation.


What do you think is true of the square roots of a complex number?

I posted an answer about cube roots of complex numbers. The same info can be applied to square roots. (see related links)


Does neutralisation in plants occur in the roots?

yes it does occur in the roots as it takes nutrients from the soil.


How can a rational equation have more than one solution?

A rational equation can be multiplied by the least common multiple of its denominators to make it into a polynomial equation. The degree of this polynomial is the highest power (of the variable) that appears in it. It can be proven that a polynomial of degree n must have n roots in the complex domain. However, there may be fewer roots in the real domain. This is because if the coefficients are real then there may be pairs of complex roots [conjugates] which will not count as real roots. Also, there may be identical roots of multiple order. For example, x4 - 1 = 0 has 4 complex roots. These are 1, -1, i and -i where i is the imaginary root of -1. There are only 2 real roots -1 and +1. x4 = 0 has 4 multiple roots, each of which is 0. Thus x = 0 is a root of multiplicity 4.


How do you simplify fourth roots?

The answer will depend on the form of the fourth root. Positive real numbers will have two fourth roots which are real and two that are complex. Complex numbers will have four complex roots. However, none of these can be "simplified" in the normal sense of the term.


What is five n squared plus two n plus six equals zero?

5n2 + 2n + 6 = 0 is a quadratic equation in the variable n.The equation does not have any real roots. The roots are the complex conjugate pair -0.2 ± 1.077i where i is the imaginary square root of -1.


If the discriminant is negative the equation has?

No real roots but the roots are a pair of complex conjugates.


What does the discriminant tell you when solving quadratic equations for the roots?

Whether the equation has 2 distinct roots, repeated roots, or complex roots. If the determinant is smaller than 0 then it has complex roots. If the determinant is 0 then it has repeated roots. If the determinant is greater than 0 then it has two distinct roots.


Could you have a quadratic function with one real root and one complex root What might the function itself look like?

Short answer: No.Assuming that the original quadratic is completely real, complex roots always come in conjugate pairs - meaning that if you multiplied both of the complex roots together, you would get a real number. For example, if one root was 2 + 3i, then you know that another root will be 2 - 3i, because those two multiplied together give you -5 (thanks to (x2 - y2) = (x+y)(x-y). You see how math all fits together? It's great!)Therefore, a real quadratic can only have 2 real roots or 2 complex roots. If you have one of each, either something has gone horribly wrong or your teacher is a sadist.Also, bear in mind that I've only done A level (American translation: late high school/early college) math, so this might be wrong if you're past that level.


How many pairs of spinal nerves are there and how do they arise?

31 pairs. And they arise from the combination of the ventral and dorsal roots.


What is the radix of x2-10x plus 31 equals 0 is x equals 5 and x equals 8?

No.The roots are the complex conjugate pair 5 ± 2.4495iwhere i is the imaginary square root of -1.

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