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

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

Solve for x (complex solution)

$$x=\sqrt{5}-2\approx 0.236067977$$

$$x=-\left(\sqrt{5}+2\right)\approx -4.236067977$$

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 -x+1.

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

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

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from 2 to get 1.

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

Subtract 2x from both sides.

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

Combine -2x and -2x to get -4x.

$$-4x+1=x^{2}$$

Subtract x^{2} from both sides.

$$-4x+1-x^{2}=0$$

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.

$$-x^{2}-4x+1=0$$

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

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

Add 16 to 4.

$$x=\frac{-\left(-4\right)±\sqrt{20}}{2\left(-1\right)}$$

Take the square root of 20.

$$x=\frac{-\left(-4\right)±2\sqrt{5}}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{5}}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{5}}{-2} when ± is plus. Add 4 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+4}{-2}$$

Divide 4+2\sqrt{5} by -2.

$$x=-\left(\sqrt{5}+2\right)$$

Now solve the equation x=\frac{4±2\sqrt{5}}{-2} when ± is minus. Subtract 2\sqrt{5} from 4.

$$x=\frac{4-2\sqrt{5}}{-2}$$

Divide 4-2\sqrt{5} by -2.

$$x=\sqrt{5}-2$$

The equation is now solved.

$$x=-\left(\sqrt{5}+2\right) x=\sqrt{5}-2$$

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 -x+1.

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

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

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

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

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

Subtract 2x from both sides.

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

Combine -2x and -2x to get -4x.

$$-4x+2=1+x^{2}$$

Subtract x^{2} from both sides.

$$-4x+2-x^{2}=1$$

Subtract 2 from both sides.

$$-4x-x^{2}=1-2$$

Subtract 2 from 1 to get -1.

$$-4x-x^{2}=-1$$

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.

$$-x^{2}-4x=-1$$

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{1}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -4 by -1.

$$x^{2}+4x=-\frac{1}{-1}$$

Divide -1 by -1.

$$x^{2}+4x=1$$

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

$$x^{2}+4x+2^{2}=1+2^{2}$$

Square 2.

$$x^{2}+4x+4=1+4$$

Add 1 to 4.

$$x^{2}+4x+4=5$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+2=\sqrt{5} x+2=-\sqrt{5}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{5}-2 x=-\sqrt{5}-2$$

Solve for x

$$x=\sqrt{5}-2\approx 0.236067977$$

$$x=-\sqrt{5}-2\approx -4.236067977$$

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 -x+1.

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

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

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from 2 to get 1.

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

Subtract 2x from both sides.

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

Combine -2x and -2x to get -4x.

$$-4x+1=x^{2}$$

Subtract x^{2} from both sides.

$$-4x+1-x^{2}=0$$

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.

$$-x^{2}-4x+1=0$$

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

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

Add 16 to 4.

$$x=\frac{-\left(-4\right)±\sqrt{20}}{2\left(-1\right)}$$

Take the square root of 20.

$$x=\frac{-\left(-4\right)±2\sqrt{5}}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{5}}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{5}}{-2} when ± is plus. Add 4 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+4}{-2}$$

Divide 4+2\sqrt{5} by -2.

$$x=-\left(\sqrt{5}+2\right)$$

Now solve the equation x=\frac{4±2\sqrt{5}}{-2} when ± is minus. Subtract 2\sqrt{5} from 4.

$$x=\frac{4-2\sqrt{5}}{-2}$$

Divide 4-2\sqrt{5} by -2.

$$x=\sqrt{5}-2$$

The equation is now solved.

$$x=-\left(\sqrt{5}+2\right) x=\sqrt{5}-2$$

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 -x+1.

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

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

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

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

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

Subtract 2x from both sides.

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

Combine -2x and -2x to get -4x.

$$-4x+2=1+x^{2}$$

Subtract x^{2} from both sides.

$$-4x+2-x^{2}=1$$

Subtract 2 from both sides.

$$-4x-x^{2}=1-2$$

Subtract 2 from 1 to get -1.

$$-4x-x^{2}=-1$$

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.

$$-x^{2}-4x=-1$$

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{1}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -4 by -1.

$$x^{2}+4x=-\frac{1}{-1}$$

Divide -1 by -1.

$$x^{2}+4x=1$$

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

$$x^{2}+4x+2^{2}=1+2^{2}$$

Square 2.

$$x^{2}+4x+4=1+4$$

Add 1 to 4.

$$x^{2}+4x+4=5$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+2=\sqrt{5} x+2=-\sqrt{5}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{5}-2 x=-\sqrt{5}-2$$