cubic equations?

Question

can anyone give me separate cubic equations with:
1) 1 real root
2) 2 real roots
3) 1 complex root

Answer 1

(1) A cubic equation with one real root is actually easy to construct.

(x + 1)^3 = 0 has one real root, or, if you want the entire cubic equation,

x^3 + 3x^2 + 3x + 1 = 0

The real root is x = -1.

(2) Two real roots.

x(x+1)^2 = 0 would have two real roots, x = {0, -1} The full cubic equation would be

x(x^2 + 2x + 1) = 0
x^3 + 2x^2 + x = 0

(3) One complex root. The key thing to know about complex roots is that *they always come in conjugate pairs*. This means that you can't one complex root without having its conjugate pair with it. So if you mean *exactly* one complex root, then it is impossible. If you mean at least one, then the solution is below.

x(x - i)(x + i) would give us the cubic equation that we want.
x(x^2 + 1) = 0

x^3 + x = 0
x = {i, -i, 0}

Answer 2

can anyone give me separate cubic equations with:
1) 1 real root
2) 2 real roots
3) 1 complex root

1) x³ - 1 = 0

2) and 3) are only possible if the coefficients of the cubic are complex

2) x³ - ix² - x + i = 0 (where i² = -1)

3) See 2)

If the coefficients of 2 and 3 are to be real then the only way you can have 2 real roots is to have 3 and the only way to have 1 complex root is to have 2 (complex conjugates of each other)

Puggy

x³ + 2x² + x = 0 has 3 real roots

x(x² + 2x + 1) = 0

x(x + 1)(x + 1) = 0

So roots are 0, -1 AND -1 (two happen to be the same BUT it still has 3 roots)

Answer 3

1) f(x) = x^3 - x^2 + x -1 = (x - 1)(x^2 + 1)
has one real root.

2) f(x) = x^3 + x^2 - x - 1 = (x - 1)(x + 1)^2
has two real roots but one of them, x = -1, is a double root. This is the only way to get exactly two real roots. One of them has to be a double root. It should be noted however, that this would be considered to be three real roots.

3) It is not possible to get exactly one complex root. For a cubic equation, you will either have no complex roots or two of them, and if you have two, they will be conjugates. The equation in 1), for example, has two complex roots and they are conjugate.

All of the above answers assume that the coefficients of the cubic equations are real numbers. Different answers may be possible if the coefficients can be complex.

Answer 4

it relatively is fairly complicated in comparision with the quatratic as quickly as.... there's an prolonged-length formulation for this... Now, the thought at the back of this formulation is making use of a few linear transformation to the equation to wind up with the style ok*X^3 + L*X + N = 0 (would be executed with all cubic equations). Then, for the ideas one proceeds making use of the roots of solidarity. relatively, we factorize x^3 - 3*a*b + a^3 + b^3, i.e. x^3 - 3*a*b + a^3 + b^3 = (x + a + b)*(x + a*o +b*o^2)*(x + b*o +a*o^2) the place o = -a million/2 + i*sqrt(3)/2 i.e a complicated 0.33 root of solidarity Now, the complicated roots of x^3 - 3*a*b + a^3 + b^3= 0 are patently: x = - (a + b), x = - (a*o +b*o^2) and x = - (b*o + a*o^2) ideas (I) Then, we evaluate: x^3 - 3*a*b + a^3 + b^3= 0 with ok*X^3 + L*X + N = 0 or equivalently X^3 + (L/ok)*X + (N/ok) = 0. From this we acquire: L/ok = - 3*a*b and N/ok = a^3 + b^3. We proceed by making use of adjusting this simulatneous equation to locate a, b in terms of ok, L, M and substitute to ideas (I) to get some new ideas. to locate the final ideas of our preliminary equation we prepare the inverse preliminary linear tranformation to the ideas.... and voila the final cubic equation is solved... As you recognize fixing your equation calls for lots lof of algebraic calculations . . . it relatively is why the cubic formulation is a lot lots bigger than the quatratic...

Answer 5

(x + 1)^3 = x^3 + 3x^2 + 3x + 1

(x - 1)(x + 1)^2 =
(x + 1)(x^2 + 2x + 1) =
x^3 + 2x^2 + x - x^2 - 2x - 1 =
x^3 + x^2 - x - 1

(x + 1 + i)(x + 1)^2 =
(x + 1 + i)(x^2 + 2x + 1) =
x^3 + 2x^2 + x + x^2 + 2x + 1 + ix^2 + i2x + i =
x^3 + (3 + i)x^2 + (3 + i2)x + (1 + i)

Answer 6

1) x^3+x=0; 2) and 3) impossible!

Answer 7

g e e k :D