Solve 6/x^2+2x+1=1

$$\frac { 6 } { x ^ { 2 } + 2 x + 1 } = 1$$

Solve for x (complex solution)

$$x=\sqrt{6}-1\approx 1.449489743$$

$$x=-\left(\sqrt{6}+1\right)\approx -3.449489743$$

Steps Using the Quadratic Formula

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

$$6=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$6=x^{2}+2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+2x+1=6$$

Subtract 6 from both sides.

$$x^{2}+2x+1-6=0$$

Subtract 6 from 1 to get -5.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 2.

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

Multiply -4 times -5.

$$x=\frac{-2±\sqrt{4+20}}{2}$$

Add 4 to 20.

$$x=\frac{-2±\sqrt{24}}{2}$$

Take the square root of 24.

$$x=\frac{-2±2\sqrt{6}}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{6}}{2} when ± is plus. Add -2 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-2}{2}$$

Divide -2+2\sqrt{6} by 2.

$$x=\sqrt{6}-1$$

Now solve the equation x=\frac{-2±2\sqrt{6}}{2} when ± is minus. Subtract 2\sqrt{6} from -2.

$$x=\frac{-2\sqrt{6}-2}{2}$$

Divide -2-2\sqrt{6} by 2.

$$x=-\sqrt{6}-1$$

The equation is now solved.

$$x=\sqrt{6}-1 x=-\sqrt{6}-1$$

Steps for Completing the Square

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

$$6=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$6=x^{2}+2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+2x+1=6$$

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}.

$$\left(x+1\right)^{2}=6$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{6}$$

Simplify.

$$x+1=\sqrt{6} x+1=-\sqrt{6}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{6}-1 x=-\sqrt{6}-1$$

Solve for x

$$x=\sqrt{6}-1\approx 1.449489743$$

$$x=-\sqrt{6}-1\approx -3.449489743$$

Steps Using the Quadratic Formula

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

$$6=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$6=x^{2}+2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+2x+1=6$$

Subtract 6 from both sides.

$$x^{2}+2x+1-6=0$$

Subtract 6 from 1 to get -5.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 2.

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

Multiply -4 times -5.

$$x=\frac{-2±\sqrt{4+20}}{2}$$

Add 4 to 20.

$$x=\frac{-2±\sqrt{24}}{2}$$

Take the square root of 24.

$$x=\frac{-2±2\sqrt{6}}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{6}}{2} when ± is plus. Add -2 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-2}{2}$$

Divide -2+2\sqrt{6} by 2.

$$x=\sqrt{6}-1$$

Now solve the equation x=\frac{-2±2\sqrt{6}}{2} when ± is minus. Subtract 2\sqrt{6} from -2.

$$x=\frac{-2\sqrt{6}-2}{2}$$

Divide -2-2\sqrt{6} by 2.

$$x=-\sqrt{6}-1$$

The equation is now solved.

$$x=\sqrt{6}-1 x=-\sqrt{6}-1$$

Steps for Completing the Square

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

$$6=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$6=x^{2}+2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+2x+1=6$$

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}.

$$\left(x+1\right)^{2}=6$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{6}$$

Simplify.

$$x+1=\sqrt{6} x+1=-\sqrt{6}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{6}-1 x=-\sqrt{6}-1$$