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

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

Solve for x

$$x=\sqrt{2}+1\approx 2.414213562$$

$$x=1-\sqrt{2}\approx -0.414213562$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2\left(x^{2}+1\right), the least common multiple of x^{2}+1,2.

$$2\left(x+1\right)+2\left(x^{2}+1\right)\left(-\frac{1}{2}\right)=0$$

Use the distributive property to multiply 2 by x+1.

$$2x+2+2\left(x^{2}+1\right)\left(-\frac{1}{2}\right)=0$$

Multiply 2 and -\frac{1}{2} to get -1.

$$2x+2-\left(x^{2}+1\right)=0$$

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

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

Subtract 1 from 2 to get 1.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -1.

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

Add 4 to 4.

$$x=\frac{-2±\sqrt{8}}{2\left(-1\right)}$$

Take the square root of 8.

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

Multiply 2 times -1.

$$x=\frac{-2±2\sqrt{2}}{-2}$$

Now solve the equation x=\frac{-2±2\sqrt{2}}{-2} when ± is plus. Add -2 to 2\sqrt{2}.

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

Divide -2+2\sqrt{2} by -2.

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

Now solve the equation x=\frac{-2±2\sqrt{2}}{-2} when ± is minus. Subtract 2\sqrt{2} from -2.

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

Divide -2-2\sqrt{2} by -2.

$$x=\sqrt{2}+1$$

The equation is now solved.

$$x=1-\sqrt{2} x=\sqrt{2}+1$$

Steps for Completing the Square

Multiply both sides of the equation by 2\left(x^{2}+1\right), the least common multiple of x^{2}+1,2.

$$2\left(x+1\right)+2\left(x^{2}+1\right)\left(-\frac{1}{2}\right)=0$$

Use the distributive property to multiply 2 by x+1.

$$2x+2+2\left(x^{2}+1\right)\left(-\frac{1}{2}\right)=0$$

Multiply 2 and -\frac{1}{2} to get -1.

$$2x+2-\left(x^{2}+1\right)=0$$

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

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

Subtract 1 from 2 to get 1.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 2 by -1.

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

Divide -1 by -1.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Add 1 to 1.

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

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

$$x=\sqrt{2}+1 x=1-\sqrt{2}$$