Solve x-left(1+xright)^2=1

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{7}i-1}{2}\approx -0.5-1.322875656i$$

$$x=\frac{-1+\sqrt{7}i}{2}\approx -0.5+1.322875656i$$

Steps Using the Quadratic Formula

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

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

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

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

Combine x and -2x to get -x.

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

Subtract 1 from both sides.

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

Subtract 1 from -1 to get -2.

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

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.

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

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

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

Multiply -4 times -1.

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

Multiply 4 times -2.

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

Add 1 to -8.

$$x=\frac{-\left(-1\right)±\sqrt{-7}}{2\left(-1\right)}$$

Take the square root of -7.

$$x=\frac{-\left(-1\right)±\sqrt{7}i}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{7}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±\sqrt{7}i}{-2}$$

Now solve the equation x=\frac{1±\sqrt{7}i}{-2} when ± is plus. Add 1 to i\sqrt{7}.

$$x=\frac{1+\sqrt{7}i}{-2}$$

Divide 1+i\sqrt{7} by -2.

$$x=\frac{-\sqrt{7}i-1}{2}$$

Now solve the equation x=\frac{1±\sqrt{7}i}{-2} when ± is minus. Subtract i\sqrt{7} from 1.

$$x=\frac{-\sqrt{7}i+1}{-2}$$

Divide 1-i\sqrt{7} by -2.

$$x=\frac{-1+\sqrt{7}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{7}i-1}{2} x=\frac{-1+\sqrt{7}i}{2}$$

Steps for Completing the Square

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

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

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

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

Combine x and -2x to get -x.

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

Add 1 to both sides.

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

Add 1 and 1 to get 2.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -1 by -1.

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

Divide 2 by -1.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -2 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=-\frac{7}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{7}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{7}i}{2} x+\frac{1}{2}=-\frac{\sqrt{7}i}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{-1+\sqrt{7}i}{2} x=\frac{-\sqrt{7}i-1}{2}$$