Solve 3x-2=4/2x

$$3 x - 2 = \frac { 4 } { 2 x }$$

Solve for x

$$x = \frac{\sqrt{7} + 1}{3} \approx 1.215250437$$

$$x=\frac{1-\sqrt{7}}{3}\approx -0.54858377$$

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 2x.

$$3x\times 2x+2x\left(-2\right)=4$$

Multiply 3 and 2 to get 6.

$$6xx+2x\left(-2\right)=4$$

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

$$6x^{2}+2x\left(-2\right)=4$$

Multiply 2 and -2 to get -4.

$$6x^{2}-4x=4$$

Subtract 4 from both sides.

$$6x^{2}-4x-4=0$$

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 6\left(-4\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-4\right)±\sqrt{16-24\left(-4\right)}}{2\times 6}$$

Multiply -24 times -4.

$$x=\frac{-\left(-4\right)±\sqrt{16+96}}{2\times 6}$$

Add 16 to 96.

$$x=\frac{-\left(-4\right)±\sqrt{112}}{2\times 6}$$

Take the square root of 112.

$$x=\frac{-\left(-4\right)±4\sqrt{7}}{2\times 6}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{7}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{4±4\sqrt{7}}{12}$$

Now solve the equation x=\frac{4±4\sqrt{7}}{12} when ± is plus. Add 4 to 4\sqrt{7}.

$$x=\frac{4\sqrt{7}+4}{12}$$

Divide 4+4\sqrt{7} by 12.

$$x=\frac{\sqrt{7}+1}{3}$$

Now solve the equation x=\frac{4±4\sqrt{7}}{12} when ± is minus. Subtract 4\sqrt{7} from 4.

$$x=\frac{4-4\sqrt{7}}{12}$$

Divide 4-4\sqrt{7} by 12.

$$x=\frac{1-\sqrt{7}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{7}+1}{3} x=\frac{1-\sqrt{7}}{3}$$

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 2x.

$$3x\times 2x+2x\left(-2\right)=4$$

Multiply 3 and 2 to get 6.

$$6xx+2x\left(-2\right)=4$$

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

$$6x^{2}+2x\left(-2\right)=4$$

Multiply 2 and -2 to get -4.

$$6x^{2}-4x=4$$

Divide both sides by 6.

$$\frac{6x^{2}-4x}{6}=\frac{4}{6}$$

Dividing by 6 undoes the multiplication by 6.

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

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{2}{3}x=\frac{4}{6}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{2}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{7}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{7}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{7}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{\sqrt{7}}{3} x-\frac{1}{3}=-\frac{\sqrt{7}}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{\sqrt{7}+1}{3} x=\frac{1-\sqrt{7}}{3}$$