Solve 1/27=2x^2+2x/frac{1458{x^2}+1458/x}

$$\frac { 1 } { 27 } = \frac { 2 x ^ { 2 } + 2 x } { \frac { 1458 } { x ^ { 2 } } + \frac { 1458 } { x } }$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{3}i-3}{2}\approx -1.5-2.598076211i$$

$$x=\frac{-3+3\sqrt{3}i}{2}\approx -1.5+2.598076211i$$

$$x=3$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x^{2} and x is x^{2}. Multiply \frac{1458}{x} times \frac{x}{x}.

$$\frac{1}{27}=\frac{2x^{2}+2x}{\frac{1458}{x^{2}}+\frac{1458x}{x^{2}}}$$

Since \frac{1458}{x^{2}} and \frac{1458x}{x^{2}} have the same denominator, add them by adding their numerators.

$$\frac{1}{27}=\frac{2x^{2}+2x}{\frac{1458+1458x}{x^{2}}}$$

Variable x cannot be equal to 0 since division by zero is not defined. Divide 2x^{2}+2x by \frac{1458+1458x}{x^{2}} by multiplying 2x^{2}+2x by the reciprocal of \frac{1458+1458x}{x^{2}}.

$$\frac{1}{27}=\frac{\left(2x^{2}+2x\right)x^{2}}{1458+1458x}$$

Factor the expressions that are not already factored in \frac{\left(2x^{2}+2x\right)x^{2}}{1458+1458x}.

$$\frac{1}{27}=\frac{2x\left(x+1\right)x^{2}}{1458\left(x+1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{1}{27}=\frac{x\left(x+1\right)x^{2}}{729\left(x+1\right)}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{1}{27}=\frac{x^{3}\left(x+1\right)}{729\left(x+1\right)}$$

Use the distributive property to multiply x^{3} by x+1.

$$\frac{1}{27}=\frac{x^{4}+x^{3}}{729\left(x+1\right)}$$

Use the distributive property to multiply 729 by x+1.

$$\frac{1}{27}=\frac{x^{4}+x^{3}}{729x+729}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{x^{4}+x^{3}}{729x+729}=\frac{1}{27}$$

Subtract \frac{1}{27} from both sides.

$$\frac{x^{4}+x^{3}}{729x+729}-\frac{1}{27}=0$$

Factor 729x+729.

$$\frac{x^{4}+x^{3}}{729\left(x+1\right)}-\frac{1}{27}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 729\left(x+1\right) and 27 is 729\left(x+1\right). Multiply \frac{1}{27} times \frac{27\left(x+1\right)}{27\left(x+1\right)}.

$$\frac{x^{4}+x^{3}}{729\left(x+1\right)}-\frac{27\left(x+1\right)}{729\left(x+1\right)}=0$$

Since \frac{x^{4}+x^{3}}{729\left(x+1\right)} and \frac{27\left(x+1\right)}{729\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{4}+x^{3}-27\left(x+1\right)}{729\left(x+1\right)}=0$$

Do the multiplications in x^{4}+x^{3}-27\left(x+1\right).

$$\frac{x^{4}+x^{3}-27x-27}{729\left(x+1\right)}=0$$

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by 729\left(x+1\right).

$$x^{4}+x^{3}-27x-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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 x^{4}+x^{3}-27x-27 by x+1 to get x^{3}-27. Solve the equation where the result equals to 0.

$$x^{3}-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-27 by x-3 to get x^{2}+3x+9. Solve the equation where the result equals to 0.

$$x^{2}+3x+9=0$$

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 1 for a, 3 for b, and 9 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-27}}{2}$$

Solve the equation x^{2}+3x+9=0 when ± is plus and when ± is minus.

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

Remove the values that the variable cannot be equal to.

$$x=3$$

List all found solutions.

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

Variable x cannot be equal to -1.

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

Solve for x

$$x=3$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x^{2} and x is x^{2}. Multiply \frac{1458}{x} times \frac{x}{x}.

$$\frac{1}{27}=\frac{2x^{2}+2x}{\frac{1458}{x^{2}}+\frac{1458x}{x^{2}}}$$

Since \frac{1458}{x^{2}} and \frac{1458x}{x^{2}} have the same denominator, add them by adding their numerators.

$$\frac{1}{27}=\frac{2x^{2}+2x}{\frac{1458+1458x}{x^{2}}}$$

Variable x cannot be equal to 0 since division by zero is not defined. Divide 2x^{2}+2x by \frac{1458+1458x}{x^{2}} by multiplying 2x^{2}+2x by the reciprocal of \frac{1458+1458x}{x^{2}}.

$$\frac{1}{27}=\frac{\left(2x^{2}+2x\right)x^{2}}{1458+1458x}$$

Factor the expressions that are not already factored in \frac{\left(2x^{2}+2x\right)x^{2}}{1458+1458x}.

$$\frac{1}{27}=\frac{2x\left(x+1\right)x^{2}}{1458\left(x+1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{1}{27}=\frac{x\left(x+1\right)x^{2}}{729\left(x+1\right)}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{1}{27}=\frac{x^{3}\left(x+1\right)}{729\left(x+1\right)}$$

Use the distributive property to multiply x^{3} by x+1.

$$\frac{1}{27}=\frac{x^{4}+x^{3}}{729\left(x+1\right)}$$

Use the distributive property to multiply 729 by x+1.

$$\frac{1}{27}=\frac{x^{4}+x^{3}}{729x+729}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{x^{4}+x^{3}}{729x+729}=\frac{1}{27}$$

Subtract \frac{1}{27} from both sides.

$$\frac{x^{4}+x^{3}}{729x+729}-\frac{1}{27}=0$$

Factor 729x+729.

$$\frac{x^{4}+x^{3}}{729\left(x+1\right)}-\frac{1}{27}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 729\left(x+1\right) and 27 is 729\left(x+1\right). Multiply \frac{1}{27} times \frac{27\left(x+1\right)}{27\left(x+1\right)}.

$$\frac{x^{4}+x^{3}}{729\left(x+1\right)}-\frac{27\left(x+1\right)}{729\left(x+1\right)}=0$$

Since \frac{x^{4}+x^{3}}{729\left(x+1\right)} and \frac{27\left(x+1\right)}{729\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{4}+x^{3}-27\left(x+1\right)}{729\left(x+1\right)}=0$$

Do the multiplications in x^{4}+x^{3}-27\left(x+1\right).

$$\frac{x^{4}+x^{3}-27x-27}{729\left(x+1\right)}=0$$

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by 729\left(x+1\right).

$$x^{4}+x^{3}-27x-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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 x^{4}+x^{3}-27x-27 by x+1 to get x^{3}-27. Solve the equation where the result equals to 0.

$$x^{3}-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-27 by x-3 to get x^{2}+3x+9. Solve the equation where the result equals to 0.

$$x^{2}+3x+9=0$$

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 1 for a, 3 for b, and 9 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-27}}{2}$$

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

$$x\in \emptyset$$

Remove the values that the variable cannot be equal to.

$$x=3$$

List all found solutions.

$$x=-1 x=3$$

Variable x cannot be equal to -1.

$$x=3$$