Solve 4/x-5/x+1=1

$$\frac { 4 } { x } - \frac { 5 } { x + 1 } = 1$$

Solve for x (complex solution)

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\left(\sqrt{5}+1\right)\approx -3.236067977$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+1\right), the least common multiple of x,x+1.

$$\left(x+1\right)\times 4-x\times 5=x\left(x+1\right)$$

Use the distributive property to multiply x+1 by 4.

$$4x+4-x\times 5=x\left(x+1\right)$$

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

$$4x+4-x\times 5=x^{2}+x$$

Subtract x^{2} from both sides.

$$4x+4-x\times 5-x^{2}=x$$

Subtract x from both sides.

$$4x+4-x\times 5-x^{2}-x=0$$

Combine 4x and -x to get 3x.

$$3x+4-x\times 5-x^{2}=0$$

Multiply -1 and 5 to get -5.

$$3x+4-5x-x^{2}=0$$

Combine 3x and -5x to get -2x.

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

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.

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

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

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

Square -2.

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

Multiply -4 times -1.

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

Multiply 4 times 4.

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

Add 4 to 16.

$$x=\frac{-\left(-2\right)±\sqrt{20}}{2\left(-1\right)}$$

Take the square root of 20.

$$x=\frac{-\left(-2\right)±2\sqrt{5}}{2\left(-1\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{2±2\sqrt{5}}{-2}$$

Now solve the equation x=\frac{2±2\sqrt{5}}{-2} when ± is plus. Add 2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+2}{-2}$$

Divide 2+2\sqrt{5} by -2.

$$x=-\left(\sqrt{5}+1\right)$$

Now solve the equation x=\frac{2±2\sqrt{5}}{-2} when ± is minus. Subtract 2\sqrt{5} from 2.

$$x=\frac{2-2\sqrt{5}}{-2}$$

Divide 2-2\sqrt{5} by -2.

$$x=\sqrt{5}-1$$

The equation is now solved.

$$x=-\left(\sqrt{5}+1\right) x=\sqrt{5}-1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+1\right), the least common multiple of x,x+1.

$$\left(x+1\right)\times 4-x\times 5=x\left(x+1\right)$$

Use the distributive property to multiply x+1 by 4.

$$4x+4-x\times 5=x\left(x+1\right)$$

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

$$4x+4-x\times 5=x^{2}+x$$

Subtract x^{2} from both sides.

$$4x+4-x\times 5-x^{2}=x$$

Subtract x from both sides.

$$4x+4-x\times 5-x^{2}-x=0$$

Combine 4x and -x to get 3x.

$$3x+4-x\times 5-x^{2}=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$3x-x\times 5-x^{2}=-4$$

Multiply -1 and 5 to get -5.

$$3x-5x-x^{2}=-4$$

Combine 3x and -5x to get -2x.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -2 by -1.

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

Divide -4 by -1.

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

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.

$$x^{2}+2x+1^{2}=4+1^{2}$$

Square 1.

$$x^{2}+2x+1=4+1$$

Add 4 to 1.

$$x^{2}+2x+1=5$$

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}.

$$\left(x+1\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Solve for x

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\sqrt{5}-1\approx -3.236067977$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+1\right), the least common multiple of x,x+1.

$$\left(x+1\right)\times 4-x\times 5=x\left(x+1\right)$$

Use the distributive property to multiply x+1 by 4.

$$4x+4-x\times 5=x\left(x+1\right)$$

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

$$4x+4-x\times 5=x^{2}+x$$

Subtract x^{2} from both sides.

$$4x+4-x\times 5-x^{2}=x$$

Subtract x from both sides.

$$4x+4-x\times 5-x^{2}-x=0$$

Combine 4x and -x to get 3x.

$$3x+4-x\times 5-x^{2}=0$$

Multiply -1 and 5 to get -5.

$$3x+4-5x-x^{2}=0$$

Combine 3x and -5x to get -2x.

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

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.

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

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

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

Square -2.

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

Multiply -4 times -1.

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

Multiply 4 times 4.

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

Add 4 to 16.

$$x=\frac{-\left(-2\right)±\sqrt{20}}{2\left(-1\right)}$$

Take the square root of 20.

$$x=\frac{-\left(-2\right)±2\sqrt{5}}{2\left(-1\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{2±2\sqrt{5}}{-2}$$

Now solve the equation x=\frac{2±2\sqrt{5}}{-2} when ± is plus. Add 2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+2}{-2}$$

Divide 2+2\sqrt{5} by -2.

$$x=-\left(\sqrt{5}+1\right)$$

Now solve the equation x=\frac{2±2\sqrt{5}}{-2} when ± is minus. Subtract 2\sqrt{5} from 2.

$$x=\frac{2-2\sqrt{5}}{-2}$$

Divide 2-2\sqrt{5} by -2.

$$x=\sqrt{5}-1$$

The equation is now solved.

$$x=-\left(\sqrt{5}+1\right) x=\sqrt{5}-1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+1\right), the least common multiple of x,x+1.

$$\left(x+1\right)\times 4-x\times 5=x\left(x+1\right)$$

Use the distributive property to multiply x+1 by 4.

$$4x+4-x\times 5=x\left(x+1\right)$$

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

$$4x+4-x\times 5=x^{2}+x$$

Subtract x^{2} from both sides.

$$4x+4-x\times 5-x^{2}=x$$

Subtract x from both sides.

$$4x+4-x\times 5-x^{2}-x=0$$

Combine 4x and -x to get 3x.

$$3x+4-x\times 5-x^{2}=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$3x-x\times 5-x^{2}=-4$$

Multiply -1 and 5 to get -5.

$$3x-5x-x^{2}=-4$$

Combine 3x and -5x to get -2x.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -2 by -1.

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

Divide -4 by -1.

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

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.

$$x^{2}+2x+1^{2}=4+1^{2}$$

Square 1.

$$x^{2}+2x+1=4+1$$

Add 4 to 1.

$$x^{2}+2x+1=5$$

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}.

$$\left(x+1\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$