Solve x^2-2x(1+x)=0

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

Solve for x

$$x=-2$$

$$x=0$$

Steps Using Factoring

Multiply -1 and 2 to get -2.

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

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

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Factor out x.

$$x\left(-x-2\right)=0$$

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

$$x=0 x=-2$$

Steps Using the Quadratic Formula

Multiply -1 and 2 to get -2.

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

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

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

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

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

Take the square root of \left(-2\right)^{2}.

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

The opposite of -2 is 2.

$$x=\frac{2±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{2±2}{-2}$$

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

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

Divide 4 by -2.

$$x=-2$$

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

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

The equation is now solved.

$$x=-2 x=0$$

Steps for Completing the Square

Multiply -1 and 2 to get -2.

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

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

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -2 by -1.

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

Divide 0 by -1.

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

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.

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

Square 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=1 x+1=-1$$

Subtract 1 from both sides of the equation.

$$x=0 x=-2$$