Solve x+3/9x=5x+9

$$x+ \frac{ 3 }{ 9x } =5x+9$$

Solve for x

$$x=\frac{\sqrt{777}}{24}-\frac{9}{8}\approx 0.036446655$$

$$x=-\frac{\sqrt{777}}{24}-\frac{9}{8}\approx -2.286446655$$

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 9x.

$$9xx+3=5x\times 9x+9x\times 9$$

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

$$9x^{2}+3=5x\times 9x+9x\times 9$$

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

$$9x^{2}+3=5x^{2}\times 9+9x\times 9$$

Multiply 5 and 9 to get 45.

$$9x^{2}+3=45x^{2}+9x\times 9$$

Multiply 9 and 9 to get 81.

$$9x^{2}+3=45x^{2}+81x$$

Subtract 45x^{2} from both sides.

$$9x^{2}+3-45x^{2}=81x$$

Combine 9x^{2} and -45x^{2} to get -36x^{2}.

$$-36x^{2}+3=81x$$

Subtract 81x from both sides.

$$-36x^{2}+3-81x=0$$

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.

$$-36x^{2}-81x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -36 for a, -81 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-81\right)±\sqrt{\left(-81\right)^{2}-4\left(-36\right)\times 3}}{2\left(-36\right)}$$

Square -81.

$$x=\frac{-\left(-81\right)±\sqrt{6561-4\left(-36\right)\times 3}}{2\left(-36\right)}$$

Multiply -4 times -36.

$$x=\frac{-\left(-81\right)±\sqrt{6561+144\times 3}}{2\left(-36\right)}$$

Multiply 144 times 3.

$$x=\frac{-\left(-81\right)±\sqrt{6561+432}}{2\left(-36\right)}$$

Add 6561 to 432.

$$x=\frac{-\left(-81\right)±\sqrt{6993}}{2\left(-36\right)}$$

Take the square root of 6993.

$$x=\frac{-\left(-81\right)±3\sqrt{777}}{2\left(-36\right)}$$

The opposite of -81 is 81.

$$x=\frac{81±3\sqrt{777}}{2\left(-36\right)}$$

Multiply 2 times -36.

$$x=\frac{81±3\sqrt{777}}{-72}$$

Now solve the equation x=\frac{81±3\sqrt{777}}{-72} when ± is plus. Add 81 to 3\sqrt{777}.

$$x=\frac{3\sqrt{777}+81}{-72}$$

Divide 81+3\sqrt{777} by -72.

$$x=-\frac{\sqrt{777}}{24}-\frac{9}{8}$$

Now solve the equation x=\frac{81±3\sqrt{777}}{-72} when ± is minus. Subtract 3\sqrt{777} from 81.

$$x=\frac{81-3\sqrt{777}}{-72}$$

Divide 81-3\sqrt{777} by -72.

$$x=\frac{\sqrt{777}}{24}-\frac{9}{8}$$

The equation is now solved.

$$x=-\frac{\sqrt{777}}{24}-\frac{9}{8} x=\frac{\sqrt{777}}{24}-\frac{9}{8}$$

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 9x.

$$9xx+3=5x\times 9x+9x\times 9$$

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

$$9x^{2}+3=5x\times 9x+9x\times 9$$

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

$$9x^{2}+3=5x^{2}\times 9+9x\times 9$$

Multiply 5 and 9 to get 45.

$$9x^{2}+3=45x^{2}+9x\times 9$$

Multiply 9 and 9 to get 81.

$$9x^{2}+3=45x^{2}+81x$$

Subtract 45x^{2} from both sides.

$$9x^{2}+3-45x^{2}=81x$$

Combine 9x^{2} and -45x^{2} to get -36x^{2}.

$$-36x^{2}+3=81x$$

Subtract 81x from both sides.

$$-36x^{2}+3-81x=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$-36x^{2}-81x=-3$$

Divide both sides by -36.

$$\frac{-36x^{2}-81x}{-36}=-\frac{3}{-36}$$

Dividing by -36 undoes the multiplication by -36.

$$x^{2}+\left(-\frac{81}{-36}\right)x=-\frac{3}{-36}$$

Reduce the fraction \frac{-81}{-36} to lowest terms by extracting and canceling out 9.

$$x^{2}+\frac{9}{4}x=-\frac{3}{-36}$$

Reduce the fraction \frac{-3}{-36} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{9}{4}x=\frac{1}{12}$$

Divide \frac{9}{4}, the coefficient of the x term, by 2 to get \frac{9}{8}. Then add the square of \frac{9}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{4}x+\left(\frac{9}{8}\right)^{2}=\frac{1}{12}+\left(\frac{9}{8}\right)^{2}$$

Square \frac{9}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{4}x+\frac{81}{64}=\frac{1}{12}+\frac{81}{64}$$

Add \frac{1}{12} to \frac{81}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{4}x+\frac{81}{64}=\frac{259}{192}$$

Factor x^{2}+\frac{9}{4}x+\frac{81}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{8}\right)^{2}=\frac{259}{192}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{8}\right)^{2}}=\sqrt{\frac{259}{192}}$$

Simplify.

$$x+\frac{9}{8}=\frac{\sqrt{777}}{24} x+\frac{9}{8}=-\frac{\sqrt{777}}{24}$$

Subtract \frac{9}{8} from both sides of the equation.

$$x=\frac{\sqrt{777}}{24}-\frac{9}{8} x=-\frac{\sqrt{777}}{24}-\frac{9}{8}$$