Solve 125w^3-5=22

$$125 w ^ { 3 } - 5 = 22$$

Solve for w

$$w=\frac{3}{5}=0.6$$

Solution Steps

Subtract 22 from both sides.

$$125w^{3}-5-22=0$$

Subtract 22 from -5 to get -27.

$$125w^{3}-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 125. List all candidates \frac{p}{q}.

$$±\frac{27}{125},±\frac{27}{25},±\frac{27}{5},±27,±\frac{9}{125},±\frac{9}{25},±\frac{9}{5},±9,±\frac{3}{125},±\frac{3}{25},±\frac{3}{5},±3,±\frac{1}{125},±\frac{1}{25},±\frac{1}{5},±1$$

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.

$$w=\frac{3}{5}$$

By Factor theorem, w-k is a factor of the polynomial for each root k. Divide 125w^{3}-27 by 5\left(w-\frac{3}{5}\right)=5w-3 to get 25w^{2}+15w+9. Solve the equation where the result equals to 0.

$$25w^{2}+15w+9=0$$

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 25 for a, 15 for b, and 9 for c in the quadratic formula.

$$w=\frac{-15±\sqrt{15^{2}-4\times 25\times 9}}{2\times 25}$$

Do the calculations.

$$w=\frac{-15±\sqrt{-675}}{50}$$

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

$$w\in \emptyset$$

List all found solutions.

$$w=\frac{3}{5}$$