Solve -2x+3x^3=20

$$- 2 x + 3 x ^ { 3 } = 20$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{21}i}{3}-1\approx -1-1.527525232i$$

$$x=2$$

$$x=\frac{\sqrt{21}i}{3}-1\approx -1+1.527525232i$$

Solution Steps

Subtract 20 from both sides.

$$-2x+3x^{3}-20=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$3x^{3}-2x-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{20}{3},±20,±\frac{10}{3},±10,±\frac{5}{3},±5,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}-2x-20 by x-2 to get 3x^{2}+6x+10. Solve the equation where the result equals to 0.

$$3x^{2}+6x+10=0$$

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 3 for a, 6 for b, and 10 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 3\times 10}}{2\times 3}$$

Do the calculations.

$$x=\frac{-6±\sqrt{-84}}{6}$$

Solve the equation 3x^{2}+6x+10=0 when ± is plus and when ± is minus.

$$x=-\frac{\sqrt{21}i}{3}-1 x=\frac{\sqrt{21}i}{3}-1$$

List all found solutions.

$$x=2 x=-\frac{\sqrt{21}i}{3}-1 x=\frac{\sqrt{21}i}{3}-1$$

Solve for x

$$x=2$$

Solution Steps

Subtract 20 from both sides.

$$-2x+3x^{3}-20=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$3x^{3}-2x-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{20}{3},±20,±\frac{10}{3},±10,±\frac{5}{3},±5,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}-2x-20 by x-2 to get 3x^{2}+6x+10. Solve the equation where the result equals to 0.

$$3x^{2}+6x+10=0$$

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 3 for a, 6 for b, and 10 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 3\times 10}}{2\times 3}$$

Do the calculations.

$$x=\frac{-6±\sqrt{-84}}{6}$$

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.

$$x=2$$