The answer will depend on 1 sixth WHAT! A sixth of a day, an hour, a year?
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.
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.
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.
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.
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The discriminant. b^2 - 4ac answer > 1; two real roots answer = 1 one real root answer < 1 no real roots
The answer will depend on 1 sixth WHAT! A sixth of a day, an hour, a year?
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.
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.
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.
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.
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.
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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.
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.
3√0.125 = 0.5 (there is only 1 real cube root of 0.125).