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

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

Solve for x

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

$$x=-1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x^{2}+x+1\right), the least common multiple of 2x^{2}+x+1,x.

$$x\times 2=-\left(2x^{2}+x+1\right)$$

To find the opposite of 2x^{2}+x+1, find the opposite of each term.

$$x\times 2=-2x^{2}-x-1$$

Add 2x^{2} to both sides.

$$x\times 2+2x^{2}=-x-1$$

Add x to both sides.

$$x\times 2+2x^{2}+x=-1$$

Combine x\times 2 and x to get 3x.

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

Add 1 to both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\times 1=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite 2x^{2}+3x+1 as \left(2x^{2}+x\right)+\left(2x+1\right).

$$\left(2x^{2}+x\right)+\left(2x+1\right)$$

Factor out x in 2x^{2}+x.

$$x\left(2x+1\right)+2x+1$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(x+1\right)$$

To find equation solutions, solve 2x+1=0 and x+1=0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x^{2}+x+1\right), the least common multiple of 2x^{2}+x+1,x.

$$x\times 2=-\left(2x^{2}+x+1\right)$$

To find the opposite of 2x^{2}+x+1, find the opposite of each term.

$$x\times 2=-2x^{2}-x-1$$

Add 2x^{2} to both sides.

$$x\times 2+2x^{2}=-x-1$$

Add x to both sides.

$$x\times 2+2x^{2}+x=-1$$

Combine x\times 2 and x to get 3x.

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

Add 1 to both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2}}{2\times 2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8}}{2\times 2}$$

Add 9 to -8.

$$x=\frac{-3±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-3±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±1}{4}$$

Now solve the equation x=\frac{-3±1}{4} when ± is plus. Add -3 to 1.

$$x=-\frac{2}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-3±1}{4} when ± is minus. Subtract 1 from -3.

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

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x\left(2x^{2}+x+1\right), the least common multiple of 2x^{2}+x+1,x.

$$x\times 2=-\left(2x^{2}+x+1\right)$$

To find the opposite of 2x^{2}+x+1, find the opposite of each term.

$$x\times 2=-2x^{2}-x-1$$

Add 2x^{2} to both sides.

$$x\times 2+2x^{2}=-x-1$$

Add x to both sides.

$$x\times 2+2x^{2}+x=-1$$

Combine x\times 2 and x to get 3x.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by 2.

$$\frac{2x^{2}+3x}{2}=-\frac{1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{3}{2}x=-\frac{1}{2}$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=-\frac{1}{2}+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Add -\frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{1}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{1}{4} x+\frac{3}{4}=-\frac{1}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

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