Solve 28x^3-27x^2+9x-1=0

$$28 x ^ { 3 } - 27 x ^ { 2 } + 9 x - 1 = 0$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{3}i+5}{14}\approx 0.357142857-0.123717915i$$

$$x=\frac{1}{4}=0.25$$

$$x=\frac{5+\sqrt{3}i}{14}\approx 0.357142857+0.123717915i$$

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

$$±\frac{1}{28},±\frac{1}{14},±\frac{1}{7},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=\frac{1}{4}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 28x^{3}-27x^{2}+9x-1 by 4\left(x-\frac{1}{4}\right)=4x-1 to get 7x^{2}-5x+1. Solve the equation where the result equals to 0.

$$7x^{2}-5x+1=0$$

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 7 for a, -5 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 7\times 1}}{2\times 7}$$

Do the calculations.

$$x=\frac{5±\sqrt{-3}}{14}$$

Solve the equation 7x^{2}-5x+1=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{3}i+5}{14} x=\frac{5+\sqrt{3}i}{14}$$

List all found solutions.

$$x=\frac{1}{4} x=\frac{-\sqrt{3}i+5}{14} x=\frac{5+\sqrt{3}i}{14}$$

Solve for x

$$x=\frac{1}{4}=0.25$$

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

$$±\frac{1}{28},±\frac{1}{14},±\frac{1}{7},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=\frac{1}{4}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 28x^{3}-27x^{2}+9x-1 by 4\left(x-\frac{1}{4}\right)=4x-1 to get 7x^{2}-5x+1. Solve the equation where the result equals to 0.

$$7x^{2}-5x+1=0$$

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 7 for a, -5 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 7\times 1}}{2\times 7}$$

Do the calculations.

$$x=\frac{5±\sqrt{-3}}{14}$$

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

$$x\in \emptyset$$

List all found solutions.

$$x=\frac{1}{4}$$