Solve 1/1=R_2^3/11,86^2

$$\frac { 1 } { 1 } = \frac { R _ { 2 } ^ { 3 } } { 11,86 ^ { 2 } }$$

Solve for R_2

$$R_{2}=1$$

Solution Steps

Anything divided by one gives itself.

$$1=\frac{R_{2}^{3}\times 11\times 86^{2}}{11\times 86^{2}}$$

Cancel out 11\times 86^{2} in both numerator and denominator.

$$1=R_{2}^{3}$$

Swap sides so that all variable terms are on the left hand side.

$$R_{2}^{3}=1$$

Subtract 1 from both sides.

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

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.

$$R_{2}=1$$

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

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

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.

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

Do the calculations.

$$R_{2}=\frac{-1±\sqrt{-3}}{2}$$

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

$$R_{2}\in \emptyset$$

List all found solutions.

$$R_{2}=1$$