Solve 2x^3-x-1=0

$$2 x ^ { 3 } - x - 1 = 0$$

Solve for x (complex solution)

$$x=-\frac{1}{2}-\frac{1}{2}i=-0.5-0.5i$$

$$x=1$$

$$x=-\frac{1}{2}+\frac{1}{2}i=-0.5+0.5i$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-x-1 by x-1 to get 2x^{2}+2x+1. Solve the equation where the result equals to 0.

$$2x^{2}+2x+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, 2 for b, and 1 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\times 1}}{2\times 2}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-4}}{4}$$

Solve the equation 2x^{2}+2x+1=0 when ± is plus and when ± is minus.

$$x=-\frac{1}{2}-\frac{1}{2}i x=-\frac{1}{2}+\frac{1}{2}i$$

List all found solutions.

$$x=1 x=-\frac{1}{2}-\frac{1}{2}i x=-\frac{1}{2}+\frac{1}{2}i$$

Solve for x

$$x=1$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-x-1 by x-1 to get 2x^{2}+2x+1. Solve the equation where the result equals to 0.

$$2x^{2}+2x+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, 2 for b, and 1 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\times 1}}{2\times 2}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-4}}{4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=1$$