Guide :

# complex, rational and real roots

2x^5 - 5x+12=0  state the number of complex roots, the possible number of real roots and the possible rational roots

## Research, Knowledge and Information :

### Rational, Irrational and Complex | Where the Arts Meet the ...

Jun 02, 2009 · Rational, Irrational and Complex. ... determine any complex roots. I still teach the Rational ... real roots? How can we find the complex roots of ...

### What are rational and irrational numbers? Topics in precalculus

What is a real number? A proof that square root of 2 is irrational. ... Rational and irrational numbers. ... the 5th root of 32 is rational, ...

### Number of possible real roots of a polynomial (video) | Khan ...

Number of possible real roots of a polynomial. Next tutorial. Finding zeros of polynomials. ... let's think about what that would imply about the non-real complex roots.

### Rational, Irrational, Real, and Imaginary Numbers | Free ...

Rational, irrational, real, and imaginary ... Density of the Number Line and Real Numbers If all rational ... It has no real solution, because the square root ...

### The Rational Roots Test - Purple Math

When a zero is a real (that is, not complex) ... The Rational Roots (or Rational Zeroes) ... The Rational Roots Test only gives a list of good first guesses; ...

### SparkNotes: Complex Numbers: Complex Roots

A summary of Complex Roots in 's Complex Numbers. ... (keep in mind that a complex number can be real if the imaginary part of the complex root is zero).

### Properties of polynomial roots - Wikipedia

Properties of polynomial roots ... An upper bound for the positive real roots is given by the sum of the two ... The complex roots can be shown to be on or ...

## Suggested Questions And Answer :

### complex, rational and real roots

The quintic function should have 5 roots.  The changes of sign (through Descartes) tell us the maximum number of positive roots. Since there are two changes of sign there is a maximum of 2 positive roots. To find the number of negative roots we negate the terms with odd powers and check for sign changes: that means that there is at most one negative root, because there's only one sign change between the first and second terms. Complex roots always come in matching or complementary pairs, so that means in this case 2 or 4.  Put x=-1: function is positive; for x=-2 function is negative, so there is a real root between -1 and -2, because the x axis must be intercepted between -1 and -2. This fulfils the maximum for negative roots. That leaves 4 more roots. They could be all complex; there could be two complex and two positive roots. Therefore there are at least two complex roots. To go deeper we can look at calculus and a graph of the function: The gradient of the function is 10x^4-5. When this is zero there is a turning point: x^4=1/2. If we differentiate again we get 40x^3. When x is negative this value is negative so the turning point is a maximum when we take the negative fourth root of 1/2; at the positive fourth root of 1/2 the turning point is a minimum, and these are irrational numbers. The value of the function at these turning points is positive. The fourth root of 1/2 is about 0.84. and once the graph has crossed the x axis between -1 and -2, it stays positive, so all other roots must be complex. The function is 12 when x=0, so we can now see the behaviour: from negative values of x, the function intersects the x axis between -1 and -2 (real root); at x=4th root of 1/2 (-0.84) it reaches a local maximum, intercepts the vertical axis at 12 until it reaches 4th root of 1/2 (0.84) and a local minimum, after which it ascends rapidly at a steep gradient.  [Incidentally, one way to find the real root is to rearrange the equation: x^5=2.5x-6=-6(1-5x/12); x=-(6(1-5x/12))^(1/5)=-6^(1/5)(1-5x/12)^(1/5) We can now use an iterative process to find x. We start with x=0, so x0, the first approximation of the solution, is the fifth root of 6 negated=-1.430969 approx. To find the next solution x1, we put x=x0 and repeat the process to get -1.571266. We keep repeating the process until we get the accuracy we need, or the calculator reaches a fixed value. After just a few iterations, my calculator gave me -1.583590704.]

### find a third degree polynomial equation with rational coefficients that has roots -4 and 6+i

A cubic equation always has three roots. These three roots are: 1) three real roots or, 2) 1 real root and two complex roots If one of the two complex roots is a + ib, then the other complex root is a - ib. We are given two of the roots as -4 and 6+i. Since one of the roots is complex and equals 6+i, the the other complex root is 6-i. Our three roots then are: -4, 6+i, 6-i. Our three solutions to the cubic equation are: x = -4, x = 6+i, x = 6-i. Which can be rewritten as: x + 4 = 0, x - (6+i) = 0, x - (6-i) = 0 Multiplying these together gives us the original cubic equation. (x + 4)(x - (6+i))(x - (6-i)) = 0 Multiplying this out, (x + 4)(x^2 - (6-i)x - (6+i)x + (6-i)(6+i)) = 0 (x + 4)(x^2 - 12x + 6^2 - i^2) = 0 (x + 4)(x^2 - 12x + 37) = 0 x^3 - 8x^2 - 11x +148 = 0

### question related to sequence and series with relation with roots of equation

a(x-x1)(x-x2)=ax^2-a(x1+x2)x+ax1x2=ax^2+bx+c=0. b=-a(x1+x2), c=ax1x2 and b^2-4ac=a^2(x1+x2)^2-4a^2x1x2=a^2(x1-x2)^2. Similarly, q=-p(x3+x4), c=px3x4 and p^2-4pr=p^2(x3-x4)^2, so (b^2-4ac)/(q^2-4pr)=(a^2(x1-x2)^2)/(p^2(x3-x4)^2) or ((a(x1-x2))/(p(x3-x4)))^2, where x3 and x4 are roots of the second polynomial. If x3 and x4 are replaced by x^3 and x^4 then the division becomes ((a(x1-x2))/(p(x^3-x^4)))^2. (i) If AP is x1, x2, 1/x3, 1/x4, then if d is the common difference, x2=x1+d so x1-x2=-d or (x1-x2)^2=d^2. If 1/x4-1x3=d, then x3-x4=dx3x4, x4(1+dx3)=x3, x4=x3/(1+dx3). Also, if 1/x3-x2=d, then 1-x2x3=dx3 and x3(d+x2)=1, x3=1/(d+x2). Let A=x1 then x2=A+d, 1/x3=A+2d, 1/x4=A+3d. x3=1/(A+2d), x4=1/(A+3d). x3-x4=d/((A+3d)(A+2d)). The division becomes: (a(A+2d)(A+3d)/p)^2=(a(x1+2d)(x1+3d)/p)^2.  (ii) If a, b, c, form a GP, then b=ar, c=ar^2, where r=common ratio. b^2-4ac=a^2r^2-4a^2r^2=-3a^2r^2. Since square root of b^2-4ac must be greater than or equal to zero for real roots of the quadratic to exist, if a, b and c are consecutive terms in a GP, then there can be no real roots. The imaginary part of the root is given by +arisqrt(3)/2a=+risqrt(3)/2 where i is the imaginary square root of -1, and x1 and x2 are both complex numbers. If x1, x2, x3, x4 form a GP with common ratio R, x1=A, x2=AR, x3=AR^2, x4=AR^3. q=-p(x3+x4)=-pAR^2(1+R) and r=px3x4=pA^2R^5. p, q, r form a series p, -pAR^2(1+R), pA^2R^5, which is not an obvious GP, because there is no visible common ratio. However, if we call the supposed common ration X, then R must be such that pX=-pAR^2(1+R) and pX^2=pA^2R^5, making X=-pA^2R^5/pAR^2(1+R)=-AR^3/(1+R) and X=-AR^2(1+R). So R^3/(1+R)=R^2(1+R); R=(1+R)^2; R^2+R+1=0, which has no real roots. The solution is a complex number: (-1+isqrt(3))/2 or (-1-isqrt(3))/2. From this, X=-A(1-3+2isqrt(3))/2isqrt(3)=x1(1+isqrt(3))/isqrt(3)=-x1(isqrt(3)/3+1). This common ratio applies only to p, q and r.

### Finding a polynomial of a given degree with given zeros: Complex zeros

If the polynomial has an even degree, meaning that the highest power of the variable (for example: x) is an even number (for example: x^4), then all the zeroes could be complex. If odd, there must be at least one real zero. A complex zero is given by the complex expression: a+ib, where a and b are real, and i=sqrt(-1). If the polynomial has real coefficients (all the numbers in at are, or represent, real numbers), as is usually the case, then the imaginary part of the complex zeroes must combine to make a real number. For example, if one complex zero is identified as a+ib, then there will be another zero a-ib such that (x-a-ib)(x-a+ib)=x^2-2ax+a^2+b^2. (Note that, although a and b are real, they are not necessarily rational numbers.) There are no imaginary components in this expansion. Example: x^4+x^3-x-1 is degree 4 polynomial. It has real zeroes 1 and -1 so that it factorises to (x-1)(x+1)(x^2+x+1). The zeroes of x^2+x+1 are found by setting the quadratic to zero and solving using the formula; so x=(-1+sqrt(1-4))/2=(-1+sqrt(-3))/2=(-1+isqrt(3))/2. The two complex roots are -1/2-isqrt(3)/2 and -1/2+isqrt(3)/2. In this case a=-1/2 and b=sqrt(3)/2. Note that a^2+b^2=1. Example: you're told that a degree 5 polynomial has a real zero 1, and two complex zeroes 1+i and -2-i. Find all zeroes and the polynomial. The two missing zeroes will be 1-i and -2+i. The polynomial is (x-1)(x-1-i)(x-1+i)(x+2+i)(x+2-i)=(x-1)(x^2-2x+2)(x^2+4x+5)=x^5+x^4-3x^3-x^2+12x-10. Strictly speaking, the polynomial is a(x^5+x^4-3x^3-x^2+12x-10), where a is a real number, because multiplying a polynomial by a constant doesn't affect its zeroes.

### how do you solve x^3+3x^2=2x+75

x^3 + 3x^2 - 2x - 75 = 0 75 factors into 3 * 5 * 5. Things you can make with those factors:  3, 5, 15, 25, 75 If x = 3 then. . . 27 + 27 - 6 - 75 = not 0 If x = 5 then 125 + 75 - 10 - 75 = not 0 If x = -3 then -27 + 27 + 6 - 75 = not 0 If x = -5 then -125 + 75 + 10 - 75 = not 0 15, -15, 25, -25, 75, and -75 will be completely dominated by the x^3, so they won't = 0 1 and -1 would be dominated by the 75, so they won't = 0 This equation does not appear to have any rational (whole number or fraction) roots. If you put the equation in here ( http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php ) you'll see that it does have one irrational (but real) root and two complex (and irrational) roots.  I don't think, in Algebra 1, these would be something you would be expected to deal with.  You're not far enough long in math to mess with things like complex math (math using square roots of negative numbers) yet.

### In algebra, why are all odd powers reversible?

Strictly speaking "NO" power operation (raising to an exponent) is reversible, because the inverse operation is ALWAYS multi-valued, unless we restrict the domain. In the case of even powers, for example: (+3) ^2 = 9 (- 3) ^2 = 9 Therefore the reverse operation thus has two possible solutions: Sqrt(9) = +3 Sqrt(9) =  -3 Also, it is "NOT" true that odd powers are reversible, if we consider complex numbers: (+3                        ) ^3  = 27 (-1.5 + j 2.59808) ^3  = 27 (-1.5 -  j 2.59808) ^3  = 27 The cube root of three (above) therefore has a total of three solutions, one "real"  and two complex conjugates. In the general case, the inverse operation of: y = (x)^n        --->   x = nth root of (y) has "n" solution. Odd roots(y) have a single real solution and (n-1)/2 pairs of complex conjugate solutions. Even roots of (y) have two real solutions (+/-) and (n-2)/2 pairs of complex conjugate solutions, assuming that "y" is positive. In the general case, the "nth root" of any real or complex power can be plotted as a set of equally spaced vectors in the complex plane, each with a magnitude of:    (|y|  / n) separated by angles of:   (2 pi / n)   or    (360 degrees /n)   In summary:     Even powers are reversible if we restrict our domain to positive numbers.     Even powers are NOT reversible in the INTEGER or larger domains,     because the roots are multi valued.     Odd powers are reversible in the REAL domain,     but NOT in the COMPLEX or larger domains. Hope this helps, MJS

### find a polynomial function of the lowest degree with rational coefficients that has the given numbers as some of its zeros -3i,5

find a polynomial function of the lowest degree with rational coefficients that has the given numbers as some of its zeros -3i,5 You have a complex root, x = -3i Now the thing about complex roots is that they always come in pairs, as complex conjugates. If one complex root is (a + ib), then the other complex root is (a - ib) since you have x = -3i, then you must also have x = 3i. Your lowest polynomial will have three roots, so three factors, (x - 3i), (x + 3i) and (x - 5), giving a cubic function. The function then is. (x - 3i)(x + 3i)( x - 5) = (x^2 - (3i)^2)(x - 5 = (x^2 + 9)(x - 5) = x^3 + 9x - 5x^2 - 45 = f(x) = x^3 - 5x^2 + 9x - 45

First set of questions is all roots, or solutions, of the equations and second set is to find where the functions are zero, the rational roots only. The methods for roots and zeroes are basically the same. 1. 3x^2+10x+3=3x^2+9x+x+3=3x(x+3)+(x+3)=(3x+1)(x+3)=0; x=-1/3, -3 2. 5x^2+5x-x-1=5x(x+1)-(x+1)=(5x-1)(x+1)=0; x=1/5, -1 3. Cubic equation. Trial shows that x=-1 is a solution, so divide cubic by factor (x+1). Use synthetic division: -1 | 3 11  5 -3      | 3 -3 -8  3        3  8 -3 | 0 and  (x+1)(3x^2+8x-3)= (x+1)(3x^2+9x-x-3)=(x+1)(3x(x+3)-(x+3))=(x+1)(3x-1)(x+3)=0; x=-1, 1/3, -3 4. 3x^4-2x^2-5=(3x^2-5)(x^2+1)=0; 3x^2=5, so x=+/-sqrt(5/3) 5. Trial shows that 3x=-1 is a solution, i.e., x=-1/3. Use synthetic division: -1/3 | 3 10 30   9         | 3  -1 -3  -9           3   9 27 | 0      (3x+1)(x^2+3x+9)=0; x=-1/3 is the only real solution 6. (5x^2+7)(x^2-2)=0; x=+/-sqrt(2) 7. No real roots 8. (5x^2+2)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2) 9. (5x^2+9)(x^2-4)=(5x^2+9)(x+2)(x-2)=0; x=+/-2 10. (5x^2+3)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2) 11. f(x)=5x(x^2-1)-(x^2-1)=(5x-1)(x-1)(x+1); x=1/5, 1, -1 12. Try x=1/2, f(x)=0. So f(x)=(2x-1)(x^2-11x-1) has no more rational roots 13. f(x)=3x(x^2-1)-(x^2-1)=(3x-1)(x-1)(x+1); zeroes at x=1/3, 1, -1 14. f(x)=(x+2)(x^2+7x+3); only rational zero at x=-2 15. (x-3)(2x^2+7x-3); rational zero at x=3 16. (x-1)^2(4x-1); x=1, 1/4 17. (x+1)(2x-1)(2x+1); x=-1, +/-1/2 18. (x+1)^2(3x-1); x=-1, 1/3 19. (x-5)(3x^2-7x-75); x=5, no other rational roots 20. (x-3)(2x^2+11x-4); x=3, other roots irrational