Solve x-3/-x-2=1/-x-2

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

Solve for x (complex solution)

$$x=\sqrt{14}-4\approx -0.258342613$$

$$x=-\left(\sqrt{14}+4\right)\approx -7.741657387$$

Steps Using the Quadratic Formula

Factor -x.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{-x}{-x}.

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

Since \frac{1}{-x} and \frac{2\left(-1\right)x}{-x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 1-2\left(-1\right)x.

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

Subtract \frac{1+2x}{-x} from both sides.

$$\frac{x-3}{-x-2}-\frac{1+2x}{-x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of -x-2 and -x is x\left(x+2\right). Multiply \frac{x-3}{-x-2} times \frac{-x}{-x}. Multiply \frac{1+2x}{-x} times \frac{-\left(x+2\right)}{-\left(x+2\right)}.

$$\frac{\left(x-3\right)\left(-1\right)x}{x\left(x+2\right)}-\frac{\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)}=0$$

Since \frac{\left(x-3\right)\left(-1\right)x}{x\left(x+2\right)} and \frac{\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(x-3\right)\left(-1\right)x-\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)}=0$$

Do the multiplications in \left(x-3\right)\left(-1\right)x-\left(1+2x\right)\left(-1\right)\left(x+2\right).

$$\frac{-x^{2}+3x+x+2+2x^{2}+4x}{x\left(x+2\right)}=0$$

Combine like terms in -x^{2}+3x+x+2+2x^{2}+4x.

$$\frac{x^{2}+8x+2}{x\left(x+2\right)}=0$$

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right).

$$x^{2}+8x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-8±\sqrt{64-8}}{2}$$

Add 64 to -8.

$$x=\frac{-8±\sqrt{56}}{2}$$

Take the square root of 56.

$$x=\frac{-8±2\sqrt{14}}{2}$$

Now solve the equation x=\frac{-8±2\sqrt{14}}{2} when ± is plus. Add -8 to 2\sqrt{14}.

$$x=\frac{2\sqrt{14}-8}{2}$$

Divide -8+2\sqrt{14} by 2.

$$x=\sqrt{14}-4$$

Now solve the equation x=\frac{-8±2\sqrt{14}}{2} when ± is minus. Subtract 2\sqrt{14} from -8.

$$x=\frac{-2\sqrt{14}-8}{2}$$

Divide -8-2\sqrt{14} by 2.

$$x=-\sqrt{14}-4$$

The equation is now solved.

$$x=\sqrt{14}-4 x=-\sqrt{14}-4$$

Steps for Completing the Square

Factor -x.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{-x}{-x}.

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

Since \frac{1}{-x} and \frac{2\left(-1\right)x}{-x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 1-2\left(-1\right)x.

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

Subtract \frac{1+2x}{-x} from both sides.

$$\frac{x-3}{-x-2}-\frac{1+2x}{-x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of -x-2 and -x is x\left(x+2\right). Multiply \frac{x-3}{-x-2} times \frac{-x}{-x}. Multiply \frac{1+2x}{-x} times \frac{-\left(x+2\right)}{-\left(x+2\right)}.

$$\frac{\left(x-3\right)\left(-1\right)x}{x\left(x+2\right)}-\frac{\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)}=0$$

Since \frac{\left(x-3\right)\left(-1\right)x}{x\left(x+2\right)} and \frac{\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(x-3\right)\left(-1\right)x-\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)}=0$$

Do the multiplications in \left(x-3\right)\left(-1\right)x-\left(1+2x\right)\left(-1\right)\left(x+2\right).

$$\frac{-x^{2}+3x+x+2+2x^{2}+4x}{x\left(x+2\right)}=0$$

Combine like terms in -x^{2}+3x+x+2+2x^{2}+4x.

$$\frac{x^{2}+8x+2}{x\left(x+2\right)}=0$$

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right).

$$x^{2}+8x+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+8x=-2$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=-2+4^{2}$$

Square 4.

$$x^{2}+8x+16=-2+16$$

Add -2 to 16.

$$x^{2}+8x+16=14$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=14$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{14}$$

Simplify.

$$x+4=\sqrt{14} x+4=-\sqrt{14}$$

Subtract 4 from both sides of the equation.

$$x=\sqrt{14}-4 x=-\sqrt{14}-4$$

Solve for x

$$x=\sqrt{14}-4\approx -0.258342613$$

$$x=-\sqrt{14}-4\approx -7.741657387$$

Steps Using the Quadratic Formula

Factor -x.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{-x}{-x}.

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

Since \frac{1}{-x} and \frac{2\left(-1\right)x}{-x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 1-2\left(-1\right)x.

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

Subtract \frac{1+2x}{-x} from both sides.

$$\frac{x-3}{-x-2}-\frac{1+2x}{-x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of -x-2 and -x is x\left(x+2\right). Multiply \frac{x-3}{-x-2} times \frac{-x}{-x}. Multiply \frac{1+2x}{-x} times \frac{-\left(x+2\right)}{-\left(x+2\right)}.

$$\frac{\left(x-3\right)\left(-1\right)x}{x\left(x+2\right)}-\frac{\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)}=0$$

Since \frac{\left(x-3\right)\left(-1\right)x}{x\left(x+2\right)} and \frac{\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(x-3\right)\left(-1\right)x-\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)}=0$$

Do the multiplications in \left(x-3\right)\left(-1\right)x-\left(1+2x\right)\left(-1\right)\left(x+2\right).

$$\frac{-x^{2}+3x+x+2+2x^{2}+4x}{x\left(x+2\right)}=0$$

Combine like terms in -x^{2}+3x+x+2+2x^{2}+4x.

$$\frac{x^{2}+8x+2}{x\left(x+2\right)}=0$$

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right).

$$x^{2}+8x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-8±\sqrt{64-8}}{2}$$

Add 64 to -8.

$$x=\frac{-8±\sqrt{56}}{2}$$

Take the square root of 56.

$$x=\frac{-8±2\sqrt{14}}{2}$$

Now solve the equation x=\frac{-8±2\sqrt{14}}{2} when ± is plus. Add -8 to 2\sqrt{14}.

$$x=\frac{2\sqrt{14}-8}{2}$$

Divide -8+2\sqrt{14} by 2.

$$x=\sqrt{14}-4$$

Now solve the equation x=\frac{-8±2\sqrt{14}}{2} when ± is minus. Subtract 2\sqrt{14} from -8.

$$x=\frac{-2\sqrt{14}-8}{2}$$

Divide -8-2\sqrt{14} by 2.

$$x=-\sqrt{14}-4$$

The equation is now solved.

$$x=\sqrt{14}-4 x=-\sqrt{14}-4$$

Steps for Completing the Square

Factor -x.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{-x}{-x}.

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

Since \frac{1}{-x} and \frac{2\left(-1\right)x}{-x} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 1-2\left(-1\right)x.

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

Subtract \frac{1+2x}{-x} from both sides.

$$\frac{x-3}{-x-2}-\frac{1+2x}{-x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of -x-2 and -x is x\left(x+2\right). Multiply \frac{x-3}{-x-2} times \frac{-x}{-x}. Multiply \frac{1+2x}{-x} times \frac{-\left(x+2\right)}{-\left(x+2\right)}.

$$\frac{\left(x-3\right)\left(-1\right)x}{x\left(x+2\right)}-\frac{\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)}=0$$

Since \frac{\left(x-3\right)\left(-1\right)x}{x\left(x+2\right)} and \frac{\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(x-3\right)\left(-1\right)x-\left(1+2x\right)\left(-1\right)\left(x+2\right)}{x\left(x+2\right)}=0$$

Do the multiplications in \left(x-3\right)\left(-1\right)x-\left(1+2x\right)\left(-1\right)\left(x+2\right).

$$\frac{-x^{2}+3x+x+2+2x^{2}+4x}{x\left(x+2\right)}=0$$

Combine like terms in -x^{2}+3x+x+2+2x^{2}+4x.

$$\frac{x^{2}+8x+2}{x\left(x+2\right)}=0$$

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right).

$$x^{2}+8x+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+8x=-2$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=-2+4^{2}$$

Square 4.

$$x^{2}+8x+16=-2+16$$

Add -2 to 16.

$$x^{2}+8x+16=14$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=14$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{14}$$

Simplify.

$$x+4=\sqrt{14} x+4=-\sqrt{14}$$

Subtract 4 from both sides of the equation.

$$x=\sqrt{14}-4 x=-\sqrt{14}-4$$