Solve x^3-4x=3

$${ x }^{ 3 } -4x = 3$$

Solve for x

$$x=\frac{1-\sqrt{13}}{2}\approx -1.302775638$$

$$x=-1$$

$$x = \frac{\sqrt{13} + 1}{2} \approx 2.302775638$$

Solution Steps

Subtract 3 from both sides.

$$x^{3}-4x-3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -3 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±3,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-4x-3 by x+1 to get x^{2}-x-3. Solve the equation where the result equals to 0.

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

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 1 for a, -1 for b, and -3 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{13}}{2}$$

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

$$x=\frac{1-\sqrt{13}}{2} x=\frac{\sqrt{13}+1}{2}$$

List all found solutions.

$$x=-1 x=\frac{1-\sqrt{13}}{2} x=\frac{\sqrt{13}+1}{2}$$