Solve 1/t+1/t-1=2

$$\frac { 1 } { t } + \frac { 1 } { t - 1 } = 2$$

Solve for t

$$t=\frac{\sqrt{2}}{2}+1\approx 1.707106781$$

$$t=-\frac{\sqrt{2}}{2}+1\approx 0.292893219$$

Steps Using the Quadratic Formula

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

$$t-1+t=2t\left(t-1\right)$$

Combine t and t to get 2t.

$$2t-1=2t\left(t-1\right)$$

Use the distributive property to multiply 2t by t-1.

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

Subtract 2t^{2} from both sides.

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

Add 2t to both sides.

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

Combine 2t and 2t to get 4t.

$$4t-1-2t^{2}=0$$

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.

$$-2t^{2}+4t-1=0$$

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

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

Square 4.

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

Multiply -4 times -2.

$$t=\frac{-4±\sqrt{16+8\left(-1\right)}}{2\left(-2\right)}$$

Multiply 8 times -1.

$$t=\frac{-4±\sqrt{16-8}}{2\left(-2\right)}$$

Add 16 to -8.

$$t=\frac{-4±\sqrt{8}}{2\left(-2\right)}$$

Take the square root of 8.

$$t=\frac{-4±2\sqrt{2}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$t=\frac{-4±2\sqrt{2}}{-4}$$

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

$$t=\frac{2\sqrt{2}-4}{-4}$$

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

$$t=-\frac{\sqrt{2}}{2}+1$$

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

$$t=\frac{-2\sqrt{2}-4}{-4}$$

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

$$t=\frac{\sqrt{2}}{2}+1$$

The equation is now solved.

$$t=-\frac{\sqrt{2}}{2}+1 t=\frac{\sqrt{2}}{2}+1$$

Steps for Completing the Square

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

$$t-1+t=2t\left(t-1\right)$$

Combine t and t to get 2t.

$$2t-1=2t\left(t-1\right)$$

Use the distributive property to multiply 2t by t-1.

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

Subtract 2t^{2} from both sides.

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

Add 2t to both sides.

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

Combine 2t and 2t to get 4t.

$$4t-1-2t^{2}=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$4t-2t^{2}=1$$

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.

$$-2t^{2}+4t=1$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 4 by -2.

$$t^{2}-2t=\frac{1}{-2}$$

Divide 1 by -2.

$$t^{2}-2t=-\frac{1}{2}$$

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.

$$t^{2}-2t+1=-\frac{1}{2}+1$$

Add -\frac{1}{2} to 1.

$$t^{2}-2t+1=\frac{1}{2}$$

Factor t^{2}-2t+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}.

$$\left(t-1\right)^{2}=\frac{1}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-1\right)^{2}}=\sqrt{\frac{1}{2}}$$

Simplify.

$$t-1=\frac{\sqrt{2}}{2} t-1=-\frac{\sqrt{2}}{2}$$

Add 1 to both sides of the equation.

$$t=\frac{\sqrt{2}}{2}+1 t=-\frac{\sqrt{2}}{2}+1$$