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

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

Solve for x (complex solution)

$$x=\sqrt{10}-2\approx 1.16227766$$

$$x=-\left(\sqrt{10}+2\right)\approx -5.16227766$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+3\right), the least common multiple of x,x+3.

$$\left(x+3\right)\times 2+x\left(-3\right)=x\left(x+3\right)$$

Use the distributive property to multiply x+3 by 2.

$$2x+6+x\left(-3\right)=x\left(x+3\right)$$

Combine 2x and x\left(-3\right) to get -x.

$$-x+6=x\left(x+3\right)$$

Use the distributive property to multiply x by x+3.

$$-x+6=x^{2}+3x$$

Subtract x^{2} from both sides.

$$-x+6-x^{2}=3x$$

Subtract 3x from both sides.

$$-x+6-x^{2}-3x=0$$

Combine -x and -3x to get -4x.

$$-4x+6-x^{2}=0$$

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.

$$-x^{2}-4x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$x=\frac{-\left(-4\right)±\sqrt{16+24}}{2\left(-1\right)}$$

Add 16 to 24.

$$x=\frac{-\left(-4\right)±\sqrt{40}}{2\left(-1\right)}$$

Take the square root of 40.

$$x=\frac{-\left(-4\right)±2\sqrt{10}}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{10}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{10}}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{10}}{-2} when ± is plus. Add 4 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}+4}{-2}$$

Divide 4+2\sqrt{10} by -2.

$$x=-\left(\sqrt{10}+2\right)$$

Now solve the equation x=\frac{4±2\sqrt{10}}{-2} when ± is minus. Subtract 2\sqrt{10} from 4.

$$x=\frac{4-2\sqrt{10}}{-2}$$

Divide 4-2\sqrt{10} by -2.

$$x=\sqrt{10}-2$$

The equation is now solved.

$$x=-\left(\sqrt{10}+2\right) x=\sqrt{10}-2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+3\right), the least common multiple of x,x+3.

$$\left(x+3\right)\times 2+x\left(-3\right)=x\left(x+3\right)$$

Use the distributive property to multiply x+3 by 2.

$$2x+6+x\left(-3\right)=x\left(x+3\right)$$

Combine 2x and x\left(-3\right) to get -x.

$$-x+6=x\left(x+3\right)$$

Use the distributive property to multiply x by x+3.

$$-x+6=x^{2}+3x$$

Subtract x^{2} from both sides.

$$-x+6-x^{2}=3x$$

Subtract 3x from both sides.

$$-x+6-x^{2}-3x=0$$

Combine -x and -3x to get -4x.

$$-4x+6-x^{2}=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$-4x-x^{2}=-6$$

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.

$$-x^{2}-4x=-6$$

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{4}{-1}\right)x=-\frac{6}{-1}$$

Divide -4 by -1.

$$x^{2}+4x=-\frac{6}{-1}$$

Divide -6 by -1.

$$x^{2}+4x=6$$

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.

$$x^{2}+4x+2^{2}=6+2^{2}$$

Square 2.

$$x^{2}+4x+4=6+4$$

Add 6 to 4.

$$x^{2}+4x+4=10$$

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}.

$$\left(x+2\right)^{2}=10$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{10}$$

Simplify.

$$x+2=\sqrt{10} x+2=-\sqrt{10}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{10}-2 x=-\sqrt{10}-2$$

Solve for x

$$x=\sqrt{10}-2\approx 1.16227766$$

$$x=-\sqrt{10}-2\approx -5.16227766$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+3\right), the least common multiple of x,x+3.

$$\left(x+3\right)\times 2+x\left(-3\right)=x\left(x+3\right)$$

Use the distributive property to multiply x+3 by 2.

$$2x+6+x\left(-3\right)=x\left(x+3\right)$$

Combine 2x and x\left(-3\right) to get -x.

$$-x+6=x\left(x+3\right)$$

Use the distributive property to multiply x by x+3.

$$-x+6=x^{2}+3x$$

Subtract x^{2} from both sides.

$$-x+6-x^{2}=3x$$

Subtract 3x from both sides.

$$-x+6-x^{2}-3x=0$$

Combine -x and -3x to get -4x.

$$-4x+6-x^{2}=0$$

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.

$$-x^{2}-4x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$x=\frac{-\left(-4\right)±\sqrt{16+24}}{2\left(-1\right)}$$

Add 16 to 24.

$$x=\frac{-\left(-4\right)±\sqrt{40}}{2\left(-1\right)}$$

Take the square root of 40.

$$x=\frac{-\left(-4\right)±2\sqrt{10}}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{10}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{10}}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{10}}{-2} when ± is plus. Add 4 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}+4}{-2}$$

Divide 4+2\sqrt{10} by -2.

$$x=-\left(\sqrt{10}+2\right)$$

Now solve the equation x=\frac{4±2\sqrt{10}}{-2} when ± is minus. Subtract 2\sqrt{10} from 4.

$$x=\frac{4-2\sqrt{10}}{-2}$$

Divide 4-2\sqrt{10} by -2.

$$x=\sqrt{10}-2$$

The equation is now solved.

$$x=-\left(\sqrt{10}+2\right) x=\sqrt{10}-2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+3\right), the least common multiple of x,x+3.

$$\left(x+3\right)\times 2+x\left(-3\right)=x\left(x+3\right)$$

Use the distributive property to multiply x+3 by 2.

$$2x+6+x\left(-3\right)=x\left(x+3\right)$$

Combine 2x and x\left(-3\right) to get -x.

$$-x+6=x\left(x+3\right)$$

Use the distributive property to multiply x by x+3.

$$-x+6=x^{2}+3x$$

Subtract x^{2} from both sides.

$$-x+6-x^{2}=3x$$

Subtract 3x from both sides.

$$-x+6-x^{2}-3x=0$$

Combine -x and -3x to get -4x.

$$-4x+6-x^{2}=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$-4x-x^{2}=-6$$

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.

$$-x^{2}-4x=-6$$

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{4}{-1}\right)x=-\frac{6}{-1}$$

Divide -4 by -1.

$$x^{2}+4x=-\frac{6}{-1}$$

Divide -6 by -1.

$$x^{2}+4x=6$$

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.

$$x^{2}+4x+2^{2}=6+2^{2}$$

Square 2.

$$x^{2}+4x+4=6+4$$

Add 6 to 4.

$$x^{2}+4x+4=10$$

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}.

$$\left(x+2\right)^{2}=10$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{10}$$

Simplify.

$$x+2=\sqrt{10} x+2=-\sqrt{10}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{10}-2 x=-\sqrt{10}-2$$