Solve b-2/b=2

$$b - \frac { 2 } { b } = 2$$

Solve for b

$$b=\sqrt{3}+1\approx 2.732050808$$

$$b=1-\sqrt{3}\approx -0.732050808$$

Steps Using the Quadratic Formula

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by b.

$$bb-2=2b$$

Multiply b and b to get b^{2}.

$$b^{2}-2=2b$$

Subtract 2b from both sides.

$$b^{2}-2-2b=0$$

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.

$$b^{2}-2b-2=0$$

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

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

Square -2.

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

Multiply -4 times -2.

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

Add 4 to 8.

$$b=\frac{-\left(-2\right)±\sqrt{12}}{2}$$

Take the square root of 12.

$$b=\frac{-\left(-2\right)±2\sqrt{3}}{2}$$

The opposite of -2 is 2.

$$b=\frac{2±2\sqrt{3}}{2}$$

Now solve the equation b=\frac{2±2\sqrt{3}}{2} when ± is plus. Add 2 to 2\sqrt{3}.

$$b=\frac{2\sqrt{3}+2}{2}$$

Divide 2+2\sqrt{3} by 2.

$$b=\sqrt{3}+1$$

Now solve the equation b=\frac{2±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 2.

$$b=\frac{2-2\sqrt{3}}{2}$$

Divide 2-2\sqrt{3} by 2.

$$b=1-\sqrt{3}$$

The equation is now solved.

$$b=\sqrt{3}+1 b=1-\sqrt{3}$$

Steps for Completing the Square

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by b.

$$bb-2=2b$$

Multiply b and b to get b^{2}.

$$b^{2}-2=2b$$

Subtract 2b from both sides.

$$b^{2}-2-2b=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$b^{2}-2b=2$$

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.

$$b^{2}-2b+1=2+1$$

Add 2 to 1.

$$b^{2}-2b+1=3$$

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

$$\left(b-1\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-1\right)^{2}}=\sqrt{3}$$

Simplify.

$$b-1=\sqrt{3} b-1=-\sqrt{3}$$

Add 1 to both sides of the equation.

$$b=\sqrt{3}+1 b=1-\sqrt{3}$$