Solve 81/x-81/5x=36

$$\frac { 81 } { x } - \frac { 81 } { 5 } x = 36$$

Solve for x

$$x = \frac{\sqrt{505} - 10}{9} \approx 1.385800562$$

$$x=\frac{-\sqrt{505}-10}{9}\approx -3.608022784$$

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 5x, the least common multiple of x,5.

$$5\times 81-\frac{81}{5}x\times 5x=180x$$

Multiply 5 and 81 to get 405.

$$405-\frac{81}{5}x\times 5x=180x$$

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

$$405-\frac{81}{5}x^{2}\times 5=180x$$

Cancel out 5 and 5.

$$405-81x^{2}=180x$$

Subtract 180x from both sides.

$$405-81x^{2}-180x=0$$

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.

$$-81x^{2}-180x+405=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -81 for a, -180 for b, and 405 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-180\right)±\sqrt{\left(-180\right)^{2}-4\left(-81\right)\times 405}}{2\left(-81\right)}$$

Square -180.

$$x=\frac{-\left(-180\right)±\sqrt{32400-4\left(-81\right)\times 405}}{2\left(-81\right)}$$

Multiply -4 times -81.

$$x=\frac{-\left(-180\right)±\sqrt{32400+324\times 405}}{2\left(-81\right)}$$

Multiply 324 times 405.

$$x=\frac{-\left(-180\right)±\sqrt{32400+131220}}{2\left(-81\right)}$$

Add 32400 to 131220.

$$x=\frac{-\left(-180\right)±\sqrt{163620}}{2\left(-81\right)}$$

Take the square root of 163620.

$$x=\frac{-\left(-180\right)±18\sqrt{505}}{2\left(-81\right)}$$

The opposite of -180 is 180.

$$x=\frac{180±18\sqrt{505}}{2\left(-81\right)}$$

Multiply 2 times -81.

$$x=\frac{180±18\sqrt{505}}{-162}$$

Now solve the equation x=\frac{180±18\sqrt{505}}{-162} when ± is plus. Add 180 to 18\sqrt{505}.

$$x=\frac{18\sqrt{505}+180}{-162}$$

Divide 180+18\sqrt{505} by -162.

$$x=\frac{-\sqrt{505}-10}{9}$$

Now solve the equation x=\frac{180±18\sqrt{505}}{-162} when ± is minus. Subtract 18\sqrt{505} from 180.

$$x=\frac{180-18\sqrt{505}}{-162}$$

Divide 180-18\sqrt{505} by -162.

$$x=\frac{\sqrt{505}-10}{9}$$

The equation is now solved.

$$x=\frac{-\sqrt{505}-10}{9} x=\frac{\sqrt{505}-10}{9}$$

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 5x, the least common multiple of x,5.

$$5\times 81-\frac{81}{5}x\times 5x=180x$$

Multiply 5 and 81 to get 405.

$$405-\frac{81}{5}x\times 5x=180x$$

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

$$405-\frac{81}{5}x^{2}\times 5=180x$$

Cancel out 5 and 5.

$$405-81x^{2}=180x$$

Subtract 180x from both sides.

$$405-81x^{2}-180x=0$$

Subtract 405 from both sides. Anything subtracted from zero gives its negation.

$$-81x^{2}-180x=-405$$

Divide both sides by -81.

$$\frac{-81x^{2}-180x}{-81}=-\frac{405}{-81}$$

Dividing by -81 undoes the multiplication by -81.

$$x^{2}+\left(-\frac{180}{-81}\right)x=-\frac{405}{-81}$$

Reduce the fraction \frac{-180}{-81} to lowest terms by extracting and canceling out 9.

$$x^{2}+\frac{20}{9}x=-\frac{405}{-81}$$

Divide -405 by -81.

$$x^{2}+\frac{20}{9}x=5$$

Divide \frac{20}{9}, the coefficient of the x term, by 2 to get \frac{10}{9}. Then add the square of \frac{10}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{20}{9}x+\left(\frac{10}{9}\right)^{2}=5+\left(\frac{10}{9}\right)^{2}$$

Square \frac{10}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{20}{9}x+\frac{100}{81}=5+\frac{100}{81}$$

Add 5 to \frac{100}{81}.

$$x^{2}+\frac{20}{9}x+\frac{100}{81}=\frac{505}{81}$$

Factor x^{2}+\frac{20}{9}x+\frac{100}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{10}{9}\right)^{2}=\frac{505}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{10}{9}\right)^{2}}=\sqrt{\frac{505}{81}}$$

Simplify.

$$x+\frac{10}{9}=\frac{\sqrt{505}}{9} x+\frac{10}{9}=-\frac{\sqrt{505}}{9}$$

Subtract \frac{10}{9} from both sides of the equation.

$$x=\frac{\sqrt{505}-10}{9} x=\frac{-\sqrt{505}-10}{9}$$