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

$$( 1 - x ) ( 2 + x ) = - 1$$

Solve for x

$$x = \frac{\sqrt{13} - 1}{2} \approx 1.302775638$$

$$x=\frac{-\sqrt{13}-1}{2}\approx -2.302775638$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1-x by 2+x and combine like terms.

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

Add 1 to both sides.

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

Add 2 and 1 to get 3.

$$3-x-x^{2}=0$$

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.

$$-x^{2}-x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 3}}{2\left(-1\right)}$$

Multiply 4 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1+12}}{2\left(-1\right)}$$

Add 1 to 12.

$$x=\frac{-\left(-1\right)±\sqrt{13}}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{13}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±\sqrt{13}}{-2}$$

Now solve the equation x=\frac{1±\sqrt{13}}{-2} when ± is plus. Add 1 to \sqrt{13}.

$$x=\frac{\sqrt{13}+1}{-2}$$

Divide 1+\sqrt{13} by -2.

$$x=\frac{-\sqrt{13}-1}{2}$$

Now solve the equation x=\frac{1±\sqrt{13}}{-2} when ± is minus. Subtract \sqrt{13} from 1.

$$x=\frac{1-\sqrt{13}}{-2}$$

Divide 1-\sqrt{13} by -2.

$$x=\frac{\sqrt{13}-1}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{13}-1}{2} x=\frac{\sqrt{13}-1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 1-x by 2+x and combine like terms.

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

Subtract 2 from both sides.

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

Subtract 2 from -1 to get -3.

$$-x-x^{2}=-3$$

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.

$$-x^{2}-x=-3$$

Divide both sides by -1.

$$\frac{-x^{2}-x}{-1}=-\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=-\frac{3}{-1}$$

Divide -1 by -1.

$$x^{2}+x=-\frac{3}{-1}$$

Divide -3 by -1.

$$x^{2}+x=3$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=3+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=3+\frac{1}{4}$$

Add 3 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{13}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{13}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{13}}{2} x+\frac{1}{2}=-\frac{\sqrt{13}}{2}$$

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

$$x=\frac{\sqrt{13}-1}{2} x=\frac{-\sqrt{13}-1}{2}$$