What are the roots of the equation x 3 1 0?

What are the roots of the equation x 3 1 0?

x^3 – 1 = 0 => (x-1)(x^2+x+1) = 0 => x-1=0 or x^2+x+1 = 0. So x = 1 is one of the roots. x = {-b±(√(b^2–4ac)}/2a. We get x = {-1 ± √(1^2 – 4(1)(1)}/2(1) = (-1±i√3)/2 which are imaginary.

Are the roots of the equation x 3 x 1 0?

If possible, let pq be a rational roots of x3-3x+1=0. Then, p is a factor of the constant term i.e. 1 and q is a factor of coefficient of x3 i.e. 1. Hence, the given equation has no rational roots.

How many real roots does x 3 1 have?

By the Fundamental Theorem of Algebra, you know that the equation x3=1 (or equivalently, x3−1=0) has three roots. We already know that x=1 is one of them, so you can use your precalculus knowledge of the factor theorem to factor x−1 out to give (x−1)(x2+x+1)=0.

How many roots does x 3 0 have?

4 Answers. In the case of x3=0, there is only one root.

How do you find the roots of x 3 8?

Algebra Examples

1. Replace f(x) with y . y=x3−8.
2. To find the roots of the equation, replace y with 0 and solve. 0=x3−8.
3. Rewrite the equation as x3−8=0 x 3 – 8 = 0 . x3−8=0.
4. Add 8 to both sides of the equation. x3=8.
5. Move 8 to the left side of the equation by subtracting it from both sides. x3−8=0.

How do you solve x 3 x 1?

Approximating a root using the Bisection Method : The Bisection Method is an iterative procedure to approximate a root (Root is another name for a solution of an equation).

How many roots does the cube of x 3 have?

Just as a quadratic equation may have two real roots, so a cubic equation has possibly three. But unlike a quadratic equation which may have no real solution, a cubic equation always has at least one real root.

What is the factor of x 3 1?

Rewrite 1 as 13 . Since both terms are perfect cubes, factor using the difference of cubes formula, a3−b3=(a−b)(a2+ab+b2) a 3 – b 3 = ( a – b ) ( a 2 + a b + b 2 ) where a=x and b=1 . Simplify.