Solve x/3+27/x=27/x+3

$$\frac { x } { 3 } + \frac { 27 } { x } = \frac { 27 } { x } + 3$$

Solve for x

$$x=9$$

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 3x, the least common multiple of 3,x.

$$xx+3\times 27=3\times 27+3x\times 3$$

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

$$x^{2}+3\times 27=3\times 27+3x\times 3$$

Multiply 3 and 27 to get 81.

$$x^{2}+81=3\times 27+3x\times 3$$

Multiply 3 and 27 to get 81.

$$x^{2}+81=81+3x\times 3$$

Multiply 3 and 3 to get 9.

$$x^{2}+81=81+9x$$

Subtract 81 from both sides.

$$x^{2}+81-81=9x$$

Subtract 81 from 81 to get 0.

$$x^{2}=9x$$

Subtract 9x from both sides.

$$x^{2}-9x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-9\right)^{2}.

$$x=\frac{-\left(-9\right)±9}{2}$$

The opposite of -9 is 9.

$$x=\frac{9±9}{2}$$

Now solve the equation x=\frac{9±9}{2} when ± is plus. Add 9 to 9.

$$x=\frac{18}{2}$$

Divide 18 by 2.

$$x=9$$

Now solve the equation x=\frac{9±9}{2} when ± is minus. Subtract 9 from 9.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=9 x=0$$

Variable x cannot be equal to 0.

$$x=9$$

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 3x, the least common multiple of 3,x.

$$xx+3\times 27=3\times 27+3x\times 3$$

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

$$x^{2}+3\times 27=3\times 27+3x\times 3$$

Multiply 3 and 27 to get 81.

$$x^{2}+81=3\times 27+3x\times 3$$

Multiply 3 and 27 to get 81.

$$x^{2}+81=81+3x\times 3$$

Multiply 3 and 3 to get 9.

$$x^{2}+81=81+9x$$

Subtract 9x from both sides.

$$x^{2}+81-9x=81$$

Subtract 81 from both sides.

$$x^{2}-9x=81-81$$

Subtract 81 from 81 to get 0.

$$x^{2}-9x=0$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=\frac{81}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{9}{2} x-\frac{9}{2}=-\frac{9}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=9 x=0$$

Variable x cannot be equal to 0.

$$x=9$$