Solve 4b^3-6b^2+1=0

$$4 b ^ { 3 } - 6 b ^ { 2 } + 1 = 0$$

Solve for b

$$b = \frac{\sqrt{3} + 1}{2} \approx 1.366025404$$

$$b=\frac{1-\sqrt{3}}{2}\approx -0.366025404$$

$$b=\frac{1}{2}=0.5$$

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

$$±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$b=\frac{1}{2}$$

By Factor theorem, b-k is a factor of the polynomial for each root k. Divide 4b^{3}-6b^{2}+1 by 2\left(b-\frac{1}{2}\right)=2b-1 to get 2b^{2}-2b-1. Solve the equation where the result equals to 0.

$$2b^{2}-2b-1=0$$

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

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

Do the calculations.

$$b=\frac{2±2\sqrt{3}}{4}$$

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

$$b=\frac{1-\sqrt{3}}{2} b=\frac{\sqrt{3}+1}{2}$$

List all found solutions.

$$b=\frac{1}{2} b=\frac{1-\sqrt{3}}{2} b=\frac{\sqrt{3}+1}{2}$$