Solve 3/x=1/x^2*x+4/2x*x

$$\frac { 3 } { x } = \frac { 1 } { x ^ { 2 } } \cdot x + \frac { 4 } { 2 x } \cdot x$$

Solve for x

$$x=1$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x^{2}, the least common multiple of x,x^{2},2x.

$$2x\times 3=2\times 1x+x\times 4x$$

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

$$2x\times 3=2\times 1x+x^{2}\times 4$$

Multiply 2 and 3 to get 6.

$$6x=2\times 1x+x^{2}\times 4$$

Multiply 2 and 1 to get 2.

$$6x=2x+x^{2}\times 4$$

Subtract 2x from both sides.

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

Combine 6x and -2x to get 4x.

$$4x=x^{2}\times 4$$

Subtract x^{2}\times 4 from both sides.

$$4x-x^{2}\times 4=0$$

Multiply -1 and 4 to get -4.

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

Factor out x.

$$x\left(4-4x\right)=0$$

To find equation solutions, solve x=0 and 4-4x=0.

$$x=0 x=1$$

Variable x cannot be equal to 0.

$$x=1$$

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^{2}, the least common multiple of x,x^{2},2x.

$$2x\times 3=2\times 1x+x\times 4x$$

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

$$2x\times 3=2\times 1x+x^{2}\times 4$$

Multiply 2 and 3 to get 6.

$$6x=2\times 1x+x^{2}\times 4$$

Multiply 2 and 1 to get 2.

$$6x=2x+x^{2}\times 4$$

Subtract 2x from both sides.

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

Combine 6x and -2x to get 4x.

$$4x=x^{2}\times 4$$

Subtract x^{2}\times 4 from both sides.

$$4x-x^{2}\times 4=0$$

Multiply -1 and 4 to get -4.

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

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.

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

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

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

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-4±4}{-8}$$

Now solve the equation x=\frac{-4±4}{-8} when ± is plus. Add -4 to 4.

$$x=\frac{0}{-8}$$

Divide 0 by -8.

$$x=0$$

Now solve the equation x=\frac{-4±4}{-8} when ± is minus. Subtract 4 from -4.

$$x=-\frac{8}{-8}$$

Divide -8 by -8.

$$x=1$$

The equation is now solved.

$$x=0 x=1$$

Variable x cannot be equal to 0.

$$x=1$$

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^{2}, the least common multiple of x,x^{2},2x.

$$2x\times 3=2\times 1x+x\times 4x$$

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

$$2x\times 3=2\times 1x+x^{2}\times 4$$

Multiply 2 and 3 to get 6.

$$6x=2\times 1x+x^{2}\times 4$$

Multiply 2 and 1 to get 2.

$$6x=2x+x^{2}\times 4$$

Subtract 2x from both sides.

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

Combine 6x and -2x to get 4x.

$$4x=x^{2}\times 4$$

Subtract x^{2}\times 4 from both sides.

$$4x-x^{2}\times 4=0$$

Multiply -1 and 4 to get -4.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -4.

$$\frac{-4x^{2}+4x}{-4}=\frac{0}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{4}{-4}x=\frac{0}{-4}$$

Divide 4 by -4.

$$x^{2}-x=\frac{0}{-4}$$

Divide 0 by -4.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{1}{4}$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$x=1 x=0$$

Variable x cannot be equal to 0.

$$x=1$$