Solve (x)=3x+1/x-3

$$( x ) = \frac { 3 x + 1 } { x - 3 }$$

Solve for x

$$x=\sqrt{10}+3\approx 6.16227766$$

$$x=3-\sqrt{10}\approx -0.16227766$$

Steps Using the Quadratic Formula

Subtract \frac{3x+1}{x-3} from both sides.

$$x-\frac{3x+1}{x-3}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-3}{x-3}.

$$\frac{x\left(x-3\right)}{x-3}-\frac{3x+1}{x-3}=0$$

Since \frac{x\left(x-3\right)}{x-3} and \frac{3x+1}{x-3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-3\right)-\left(3x+1\right)}{x-3}=0$$

Do the multiplications in x\left(x-3\right)-\left(3x+1\right).

$$\frac{x^{2}-3x-3x-1}{x-3}=0$$

Combine like terms in x^{2}-3x-3x-1.

$$\frac{x^{2}-6x-1}{x-3}=0$$

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

$$x^{2}-6x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-1\right)}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$x=\frac{-\left(-6\right)±\sqrt{36+4}}{2}$$

Add 36 to 4.

$$x=\frac{-\left(-6\right)±\sqrt{40}}{2}$$

Take the square root of 40.

$$x=\frac{-\left(-6\right)±2\sqrt{10}}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{10}}{2}$$

Now solve the equation x=\frac{6±2\sqrt{10}}{2} when ± is plus. Add 6 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}+6}{2}$$

Divide 6+2\sqrt{10} by 2.

$$x=\sqrt{10}+3$$

Now solve the equation x=\frac{6±2\sqrt{10}}{2} when ± is minus. Subtract 2\sqrt{10} from 6.

$$x=\frac{6-2\sqrt{10}}{2}$$

Divide 6-2\sqrt{10} by 2.

$$x=3-\sqrt{10}$$

The equation is now solved.

$$x=\sqrt{10}+3 x=3-\sqrt{10}$$

Steps for Completing the Square

Subtract \frac{3x+1}{x-3} from both sides.

$$x-\frac{3x+1}{x-3}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-3}{x-3}.

$$\frac{x\left(x-3\right)}{x-3}-\frac{3x+1}{x-3}=0$$

Since \frac{x\left(x-3\right)}{x-3} and \frac{3x+1}{x-3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-3\right)-\left(3x+1\right)}{x-3}=0$$

Do the multiplications in x\left(x-3\right)-\left(3x+1\right).

$$\frac{x^{2}-3x-3x-1}{x-3}=0$$

Combine like terms in x^{2}-3x-3x-1.

$$\frac{x^{2}-6x-1}{x-3}=0$$

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

$$x^{2}-6x-1=0$$

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

$$x^{2}-6x=1$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=1+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=1+9$$

Add 1 to 9.

$$x^{2}-6x+9=10$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=10$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{10}$$

Simplify.

$$x-3=\sqrt{10} x-3=-\sqrt{10}$$

Add 3 to both sides of the equation.

$$x=\sqrt{10}+3 x=3-\sqrt{10}$$