Solve 2x+1+1/x^2=0

$$2 x + 1 + \frac { 1 } { x ^ { 2 } } = 0$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{7}i}{4}\approx 0.25+0.661437828i$$

$$x=\frac{-\sqrt{7}i+1}{4}\approx 0.25-0.661437828i$$

$$x=-1$$

Solution Steps

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$2xx^{2}+x^{2}+1=0$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$2x^{3}+x^{2}+1=0$$

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.

$$x=-1$$

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

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

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.

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

Do the calculations.

$$x=\frac{1±\sqrt{-7}}{4}$$

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

$$x=\frac{-\sqrt{7}i+1}{4} x=\frac{1+\sqrt{7}i}{4}$$

List all found solutions.

$$x=-1 x=\frac{-\sqrt{7}i+1}{4} x=\frac{1+\sqrt{7}i}{4}$$

Solve for x

$$x=-1$$

Solution Steps

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$2xx^{2}+x^{2}+1=0$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$2x^{3}+x^{2}+1=0$$

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.

$$x=-1$$

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

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

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.

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

Do the calculations.

$$x=\frac{1±\sqrt{-7}}{4}$$

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.

$$x=-1$$