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

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

Solve for x (complex solution)

$$x=-\frac{\sqrt{3}i}{6}-\frac{1}{2}\approx -0.5-0.288675135i$$

$$x=1$$

$$x=\frac{\sqrt{3}i}{6}-\frac{1}{2}\approx -0.5+0.288675135i$$

Solution Steps

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

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x^{2} and 2x is 2x^{2}. Multiply \frac{1}{x^{2}} times \frac{2}{2}. Multiply \frac{4}{2x} times \frac{x}{x}.

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

Since \frac{2}{2x^{2}} and \frac{4x}{2x^{2}} have the same denominator, add them by adding their numerators.

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

Factor the expressions that are not already factored in \frac{2+4x}{2x^{2}}.

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

Cancel out 2 in both numerator and denominator.

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

Subtract \frac{2x+1}{x^{2}} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3x times \frac{x^{2}}{x^{2}}.

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

Since \frac{3xx^{2}}{x^{2}} and \frac{2x+1}{x^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3xx^{2}-\left(2x+1\right)}{x^{2}}=0$$

Do the multiplications in 3xx^{2}-\left(2x+1\right).

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

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{1}{3},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}-2x-1 by x-1 to get 3x^{2}+3x+1. Solve the equation where the result equals to 0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 3 for a, 3 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-3±\sqrt{-3}}{6}$$

Solve the equation 3x^{2}+3x+1=0 when ± is plus and when ± is minus.

$$x=-\frac{\sqrt{3}i}{6}-\frac{1}{2} x=\frac{\sqrt{3}i}{6}-\frac{1}{2}$$

List all found solutions.

$$x=1 x=-\frac{\sqrt{3}i}{6}-\frac{1}{2} x=\frac{\sqrt{3}i}{6}-\frac{1}{2}$$

Solve for x

$$x=1$$

Solution Steps

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

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x^{2} and 2x is 2x^{2}. Multiply \frac{1}{x^{2}} times \frac{2}{2}. Multiply \frac{4}{2x} times \frac{x}{x}.

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

Since \frac{2}{2x^{2}} and \frac{4x}{2x^{2}} have the same denominator, add them by adding their numerators.

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

Factor the expressions that are not already factored in \frac{2+4x}{2x^{2}}.

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

Cancel out 2 in both numerator and denominator.

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

Subtract \frac{2x+1}{x^{2}} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3x times \frac{x^{2}}{x^{2}}.

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

Since \frac{3xx^{2}}{x^{2}} and \frac{2x+1}{x^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3xx^{2}-\left(2x+1\right)}{x^{2}}=0$$

Do the multiplications in 3xx^{2}-\left(2x+1\right).

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

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{1}{3},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}-2x-1 by x-1 to get 3x^{2}+3x+1. Solve the equation where the result equals to 0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 3 for a, 3 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-3±\sqrt{-3}}{6}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=1$$