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How do you find a cubic equation with integral coefficients that has the roots 2 and 4 plus i?
If it has integral coefficients and 4+i is a root then its conjugate, 4-i must also be a root.
So the equation is f(x) = (x-2)*(x-4-i)*(x-4+i) where each factor is x minus a root. Then multiply these out.
= (x-2)*(x2 - 8x + 17) = x3 - 2x2 - 8x2 + 16x + 17x - 34 = x3 - 10x2 + 33x - 34
What else can I help you with?
How many real solutions does a cubic equation have?
A cubic has from 1 to 3 real solutions. The fact that every cubic equation with real coefficients has at least 1 real solution comes from the intermediate value theorem. The discriminant of the equation tells you how many roots there are.
What is the product of the roots of the equation 2xsquare-x-2 equals 0?
The product of the roots of the equation 2x2 -x -2 = 2 is 2x2 -x -2 = 2.
What is the number and type of roots of the equation x2-4x-32 equals 0?
There are 2 roots to the equation x2-4x-32 equals 0; factored it is (x-8)(x+4); therefore the roots are 8 & -4.
What are the compex roots in mathematics?
The complex roots of an equation is any solution to that equation which cannot be expressed in terms of real numbers. For example, the equation 0 = x² + 5 does not have any solution in real numbers. But in complex numbers, it has solutions.
How many real roots does the equation 2x2-5x-3 equals 0 have?
2x2 - 5x - 3 = 0 A quadratic equation expressed in the form ax2 + bx + c = 0 has two real and distinct roots when b2 - 4ac is positive. Using the figures from the supplied equation then b2 - 4ac = 52 - (4 x 2 x -3) = 25 + 24 = 49. Therefore there are TWO real and distinct roots.
How do you find a cubic equation with integral coefficients that has the roots -3 and 5 minus i?
If the equation has real coefficients, and 5 - i is a root, then its conjugate, 5 + i must be a root.Since 5 - i and 5 + i are roots, then (x - 5 + i) and (x - 5 - i) are factors.That means x2 - 10x + 26 is a factor.The other root is x = -3 so x + 3 is the other factor.So the cubic is (x + 3)*(x2 - 10x + 26) = 0That is x3 - 7x2 - 4x + 78 = 0
How many real solutions does a cubic equation have?
A cubic has from 1 to 3 real solutions. The fact that every cubic equation with real coefficients has at least 1 real solution comes from the intermediate value theorem. The discriminant of the equation tells you how many roots there are.
Does there exist a quadratic equation whose coefficients are irrational but both the roots are rational?
None, if the coefficients of the quadratic are in their lowest form.
Pi is transcendental What does this mean?
Real but not a root of an algebraic equation with rational roots coefficients
How do I find the discriminant?
Given the quadratic equation ax^2 + bx + c =0, where a, b, and c are real numbers: (The discriminant is equal to b^2 - 4ac) If b^2 - 4ac < 0, there are two conjugate imaginary roots. If b^2 - 4ac = 0, there is one real root (called double root) If b^2 - 4ac > 0, there are two different real roots. In the special case when the equation has integral coefficients (means that all coefficients are integers), and b^2 - 4ac is the square of an integer, the equation has rational roots. That is , if b^2 - 4ac is the square of an integer, then ax^2 + bx + c has factors with integral coefficients. * * * * * Strictly speaking, the last part of the last sentence is not true. For example, consider the equation 4x2 + 8x + 3 = 0 the discriminant is 16, which is a perfect square and the equation can be written as (2x+1)*(2x+3) = 0 To that extent the above is correct. However, the equation can also be written, in factorised form, as (x+1/2)*(x+3/2) = 0 Not all integral coefficients.
How many roots does a quadratic equation in one variable have?
A quadratic equation has two roots. They may be similar or dissimilar. As the highest power of a quadratic equation is 2 , there are 2 roots. Similarly, in the cubic equation, the highest power is 3, so it has three equal or unequal roots. So the highest power of an equation is the answer to the no of roots of that particular equation.
How do you factor a polynomial with the equation of 6A2 - B 2?
The given polynomial does not have factors with rational coefficients.
Roots of cubic equation?
There is a formula, but it is very difficult. I will give you a link to it. http://en.wikipedia.org/wiki/Cubic_equation
Can bisection method give us two answers for different intervals for same equation i have equation 2x3 plus x2-20x plus 12 equals 0. i have to find real root upto 3 correct decimal places. I chose int?
Yes. A cubic equation can have 3 real roots. Depending on their size, each of three intervals could contain a root. In that case different intervals must give different roots.Yes. A cubic equation can have 3 real roots. Depending on their size, each of three intervals could contain a root. In that case different intervals must give different roots.Yes. A cubic equation can have 3 real roots. Depending on their size, each of three intervals could contain a root. In that case different intervals must give different roots.Yes. A cubic equation can have 3 real roots. Depending on their size, each of three intervals could contain a root. In that case different intervals must give different roots.
If one of the roots of the quadratic equation is 1 plus 3i what is the other root?
The answer to the question, as stated, is that the other root could be anything. However, if all the coefficients of the quadratic equation are real numbers, then the other root is 1 minus 3i.
Why does the discriminant determine the number and type of the solutions for a quadratic equation?
A quadratic equation is wholly defined by its coefficients. The solutions or roots of the quadratic can, therefore, be determined by a function of these coefficients - and this function called the quadratic formula. Within this function, there is one part that specifically determines the number and types of solutions it is therefore called the discriminant: it discriminates between the different types of solutions.
How do you create a function to calculate the roots of any given quadratic equation?
First rewrite the quadratic equation in the form: ax2 + bx + c = 0 where a , b and c are constant coefficients. Clearly, a is not = 0 for if it were then you would have a linear equation and not a quadratic. Then the roots of the quadratic are: x = [-b +/- sqrt(b2 - 4ac)]/2a where using the + and - values of the square root result in two solutions.
