Solve 40/x-2=40/x+1

$$\frac { 40 } { x - 2 } = \frac { 40 } { x } + 1$$

Solve for x

$$x=-8$$

$$x=10$$

Steps Using the Quadratic Formula

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

$$x\times 40=\left(x-2\right)\times 40+x\left(x-2\right)$$

Use the distributive property to multiply x-2 by 40.

$$x\times 40=40x-80+x\left(x-2\right)$$

Use the distributive property to multiply x by x-2.

$$x\times 40=40x-80+x^{2}-2x$$

Combine 40x and -2x to get 38x.

$$x\times 40=38x-80+x^{2}$$

Subtract 38x from both sides.

$$x\times 40-38x=-80+x^{2}$$

Combine x\times 40 and -38x to get 2x.

$$2x=-80+x^{2}$$

Subtract -80 from both sides.

$$2x-\left(-80\right)=x^{2}$$

The opposite of -80 is 80.

$$2x+80=x^{2}$$

Subtract x^{2} from both sides.

$$2x+80-x^{2}=0$$

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.

$$-x^{2}+2x+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 2 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-1\right)\times 80}}{2\left(-1\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-1\right)\times 80}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-2±\sqrt{4+4\times 80}}{2\left(-1\right)}$$

Multiply 4 times 80.

$$x=\frac{-2±\sqrt{4+320}}{2\left(-1\right)}$$

Add 4 to 320.

$$x=\frac{-2±\sqrt{324}}{2\left(-1\right)}$$

Take the square root of 324.

$$x=\frac{-2±18}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-2±18}{-2}$$

Now solve the equation x=\frac{-2±18}{-2} when ± is plus. Add -2 to 18.

$$x=\frac{16}{-2}$$

Divide 16 by -2.

$$x=-8$$

Now solve the equation x=\frac{-2±18}{-2} when ± is minus. Subtract 18 from -2.

$$x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x=10$$

The equation is now solved.

$$x=-8 x=10$$

Steps for Completing the Square

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

$$x\times 40=\left(x-2\right)\times 40+x\left(x-2\right)$$

Use the distributive property to multiply x-2 by 40.

$$x\times 40=40x-80+x\left(x-2\right)$$

Use the distributive property to multiply x by x-2.

$$x\times 40=40x-80+x^{2}-2x$$

Combine 40x and -2x to get 38x.

$$x\times 40=38x-80+x^{2}$$

Subtract 38x from both sides.

$$x\times 40-38x=-80+x^{2}$$

Combine x\times 40 and -38x to get 2x.

$$2x=-80+x^{2}$$

Subtract x^{2} from both sides.

$$2x-x^{2}=-80$$

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.

$$-x^{2}+2x=-80$$

Divide both sides by -1.

$$\frac{-x^{2}+2x}{-1}=-\frac{80}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{2}{-1}x=-\frac{80}{-1}$$

Divide 2 by -1.

$$x^{2}-2x=-\frac{80}{-1}$$

Divide -80 by -1.

$$x^{2}-2x=80$$

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.

$$x^{2}-2x+1=80+1$$

Add 80 to 1.

$$x^{2}-2x+1=81$$

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}.

$$\left(x-1\right)^{2}=81$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{81}$$

Simplify.

$$x-1=9 x-1=-9$$

Add 1 to both sides of the equation.

$$x=10 x=-8$$