Solve 2alpha^3-3alpha^2+1=0

$$2 \alpha ^ { 3 } - 3 \alpha ^ { 2 } + 1 = 0$$

Solve for α

$$\alpha =1$$

$$\alpha =-\frac{1}{2}=-0.5$$

Solution Steps

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

$$±\frac{1}{2},±1$$

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.

$$\alpha =1$$

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

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

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

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

Do the calculations.

$$\alpha =\frac{1±3}{4}$$

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

$$\alpha =-\frac{1}{2} \alpha =1$$

List all found solutions.

$$\alpha =1 \alpha =-\frac{1}{2}$$