Solve 2y^3-18y^2+y-9=0

$$2 y ^ { 3 } - 18 y ^ { 2 } + y - 9 = 0$$

Solve for y (complex solution)

$$y=\frac{\sqrt{2}i}{2}\approx 0.707106781i$$

$$y=9$$

$$y=-\frac{\sqrt{2}i}{2}\approx -0-0.707106781i$$

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 -9 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{9}{2},±9,±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$y=9$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 2y^{3}-18y^{2}+y-9 by y-9 to get 2y^{2}+1. Solve the equation where the result equals to 0.

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

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 2 for a, 0 for b, and 1 for c in the quadratic formula.

$$y=\frac{0±\sqrt{0^{2}-4\times 2\times 1}}{2\times 2}$$

Do the calculations.

$$y=\frac{0±\sqrt{-8}}{4}$$

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

$$y=-\frac{\sqrt{2}i}{2} y=\frac{\sqrt{2}i}{2}$$

List all found solutions.

$$y=9 y=-\frac{\sqrt{2}i}{2} y=\frac{\sqrt{2}i}{2}$$

Solve for y

$$y=9$$

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 -9 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{9}{2},±9,±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$y=9$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 2y^{3}-18y^{2}+y-9 by y-9 to get 2y^{2}+1. Solve the equation where the result equals to 0.

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

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 2 for a, 0 for b, and 1 for c in the quadratic formula.

$$y=\frac{0±\sqrt{0^{2}-4\times 2\times 1}}{2\times 2}$$

Do the calculations.

$$y=\frac{0±\sqrt{-8}}{4}$$

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

$$y\in \emptyset$$

List all found solutions.

$$y=9$$