Solve varphi^3=2varphi+1

$$\varphi ^ { 3 } = 2 \varphi + 1$$

Solve for φ

$$\phi =\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

$$\phi =-1$$

$$\phi =\frac{\sqrt{5}+1}{2}\approx 1.618033989$$

Solution Steps

Subtract 2\phi from both sides.

$$\phi ^{3}-2\phi =1$$

Subtract 1 from both sides.

$$\phi ^{3}-2\phi -1=0$$

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 1. List all candidates \frac{p}{q}.

$$±1$$

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.

$$\phi =-1$$

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

$$\phi ^{2}-\phi -1=0$$

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

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

Do the calculations.

$$\phi =\frac{1±\sqrt{5}}{2}$$

Solve the equation \phi ^{2}-\phi -1=0 when ± is plus and when ± is minus.

$$\phi =\frac{1-\sqrt{5}}{2} \phi =\frac{\sqrt{5}+1}{2}$$

List all found solutions.

$$\phi =-1 \phi =\frac{1-\sqrt{5}}{2} \phi =\frac{\sqrt{5}+1}{2}$$