Solve u+1/u=4

$$u + \frac { 1 } { u } = 4$$

Solve for u

$$u=\sqrt{3}+2\approx 3.732050808$$

$$u=2-\sqrt{3}\approx 0.267949192$$

Steps Using the Quadratic Formula

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

$$uu+1=4u$$

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

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

Subtract 4u from both sides.

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

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.

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

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

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

Square -4.

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

Add 16 to -4.

$$u=\frac{-\left(-4\right)±\sqrt{12}}{2}$$

Take the square root of 12.

$$u=\frac{-\left(-4\right)±2\sqrt{3}}{2}$$

The opposite of -4 is 4.

$$u=\frac{4±2\sqrt{3}}{2}$$

Now solve the equation u=\frac{4±2\sqrt{3}}{2} when ± is plus. Add 4 to 2\sqrt{3}.

$$u=\frac{2\sqrt{3}+4}{2}$$

Divide 4+2\sqrt{3} by 2.

$$u=\sqrt{3}+2$$

Now solve the equation u=\frac{4±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 4.

$$u=\frac{4-2\sqrt{3}}{2}$$

Divide 4-2\sqrt{3} by 2.

$$u=2-\sqrt{3}$$

The equation is now solved.

$$u=\sqrt{3}+2 u=2-\sqrt{3}$$

Steps for Completing the Square

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

$$uu+1=4u$$

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

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

Subtract 4u from both sides.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Square -2.

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

Add -1 to 4.

$$u^{2}-4u+4=3$$

Factor u^{2}-4u+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$u-2=\sqrt{3} u-2=-\sqrt{3}$$

Add 2 to both sides of the equation.

$$u=\sqrt{3}+2 u=2-\sqrt{3}$$