Solve t+4/t=2

$$t + \frac { 4 } { t } = 2$$

Solve for t

$$t=1+\sqrt{3}i\approx 1+1.732050808i$$

$$t=-\sqrt{3}i+1\approx 1-1.732050808i$$

Steps Using the Quadratic Formula

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

$$tt+4=2t$$

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

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

Subtract 2t from both sides.

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

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.

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

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}.

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

Square -2.

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

Multiply -4 times 4.

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

Add 4 to -16.

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

Take the square root of -12.

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

The opposite of -2 is 2.

$$t=\frac{2±2\sqrt{3}i}{2}$$

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

$$t=\frac{2+2\sqrt{3}i}{2}$$

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

$$t=1+\sqrt{3}i$$

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

$$t=\frac{-2\sqrt{3}i+2}{2}$$

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

$$t=-\sqrt{3}i+1$$

The equation is now solved.

$$t=1+\sqrt{3}i t=-\sqrt{3}i+1$$

Steps for Completing the Square

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

$$tt+4=2t$$

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

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

Subtract 2t from both sides.

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

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

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

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.

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

Add -4 to 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$t-1=\sqrt{3}i t-1=-\sqrt{3}i$$

Add 1 to both sides of the equation.

$$t=1+\sqrt{3}i t=-\sqrt{3}i+1$$