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Q: How many real sixth roots does - 1 have?
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How many minutes is 1 sixth?

The answer will depend on 1 sixth WHAT! A sixth of a day, an hour, a year?


What are the types of parabolas?

There are two ways of classifying parabolas: By the direction in which they are open: open at the top or at the bottom. By the number of real roots: 2 real, 1 real or no real roots.


What is the fourth root of 1?

The two real roots are -1 and +1.There are also two roots in the complex field and these are ±i where i is the imaginary square root of -1.


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 many roots does the graph y 5x 2 - 4x 1have?

Either +/- 1, and the discriminant will tell. b2 - 4acY = 5X2 - 4 + 1(- 4)2 - 4(5)(1)= 16 - 20= - 4==========It that 1 is positive then no real roots.Y = 5x2 - 4 - 1(- 4)2 - 4(5)(- 1)16 + 20= 36==========if 1 is negative then two real roots.

Related questions

Does every sixth-degree equation have at least two real roots?

It need not have any real roots. For example x6 + 1 = 0 has none.


How Many real cube roots dose 1 have?

1


What part of a quadratic formula determines the number of x- intercepts?

The discriminant. b^2 - 4ac answer > 1; two real roots answer = 1 one real root answer < 1 no real roots


How many minutes is 1 sixth?

The answer will depend on 1 sixth WHAT! A sixth of a day, an hour, a year?


How many real cube roots does negative 1 have?

Any real number - positive or negative - has exactly one real cube root. Any real number (except zero) has three cubic roots in the complex numbers; but only one of them is real.


What are the types of parabolas?

There are two ways of classifying parabolas: By the direction in which they are open: open at the top or at the bottom. By the number of real roots: 2 real, 1 real or no real roots.


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 fourth root of 1?

The two real roots are -1 and +1.There are also two roots in the complex field and these are ±i where i is the imaginary square root of -1.


What is the sixth root of 64?

2 64 is also 26, so the sixth root of 26 is 2. In general, the n-th root of Xn is X. ALTERNATIVE ANSWER: The Fundamental Theorem of Algebra states that every polynomial of degree n has n (possibly complex) roots. This means that there are 6 sixth roots of 64. These roots are: 2*w, 2*w^2, 2*w^3, 2*w^4, 2 * w^5 and 2 * w^6 where w is the 6th root of unity, or w = e^(2πi/6) = cos(2π / 6) + i * sin (2π / 6) Note that w^3 = -1 and w^6 = 1, so there are actually two real sixth roots: 2 and -2. The other four are imaginary/complex.


How many sixth are there in 1?

6


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 many roots does the graph y 5x 2 - 4x 1have?

Either +/- 1, and the discriminant will tell. b2 - 4acY = 5X2 - 4 + 1(- 4)2 - 4(5)(1)= 16 - 20= - 4==========It that 1 is positive then no real roots.Y = 5x2 - 4 - 1(- 4)2 - 4(5)(- 1)16 + 20= 36==========if 1 is negative then two real roots.