Solve 2x-1/x=3

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

Solve for x

$$x = \frac{\sqrt{17} + 3}{4} \approx 1.780776406$$

$$x=\frac{3-\sqrt{17}}{4}\approx -0.280776406$$

Steps Using the Quadratic Formula

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

$$2xx-1=3x$$

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

$$2x^{2}-1=3x$$

Subtract 3x from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-1\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+8}}{2\times 2}$$

Add 9 to 8.

$$x=\frac{-\left(-3\right)±\sqrt{17}}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±\sqrt{17}}{4}$$

Now solve the equation x=\frac{3±\sqrt{17}}{4} when ± is plus. Add 3 to \sqrt{17}.

$$x=\frac{\sqrt{17}+3}{4}$$

Now solve the equation x=\frac{3±\sqrt{17}}{4} when ± is minus. Subtract \sqrt{17} from 3.

$$x=\frac{3-\sqrt{17}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+3}{4} x=\frac{3-\sqrt{17}}{4}$$

Steps for Completing the Square

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

$$2xx-1=3x$$

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

$$2x^{2}-1=3x$$

Subtract 3x from both sides.

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

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

$$2x^{2}-3x=1$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{2}+\frac{9}{16}$$

Add \frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{17}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{17}}{4} x-\frac{3}{4}=-\frac{\sqrt{17}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{17}+3}{4} x=\frac{3-\sqrt{17}}{4}$$