Solve 9(x)=x^2+x+1/x-2

$$9 ( x ) = \frac { x ^ { 2 } + x + 1 } { x - 2 }$$

Solve for x

$$x = \frac{\sqrt{393} + 19}{16} \approx 2.426514225$$

$$x=\frac{19-\sqrt{393}}{16}\approx -0.051514225$$

Steps Using the Quadratic Formula

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

$$9x\left(x-2\right)=x^{2}+x+1$$

Use the distributive property to multiply 9x by x-2.

$$9x^{2}-18x=x^{2}+x+1$$

Subtract x^{2} from both sides.

$$9x^{2}-18x-x^{2}=x+1$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}-18x=x+1$$

Subtract x from both sides.

$$8x^{2}-18x-x=1$$

Combine -18x and -x to get -19x.

$$8x^{2}-19x=1$$

Subtract 1 from both sides.

$$8x^{2}-19x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -19 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 8\left(-1\right)}}{2\times 8}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 8\left(-1\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-19\right)±\sqrt{361-32\left(-1\right)}}{2\times 8}$$

Multiply -32 times -1.

$$x=\frac{-\left(-19\right)±\sqrt{361+32}}{2\times 8}$$

Add 361 to 32.

$$x=\frac{-\left(-19\right)±\sqrt{393}}{2\times 8}$$

The opposite of -19 is 19.

$$x=\frac{19±\sqrt{393}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{19±\sqrt{393}}{16}$$

Now solve the equation x=\frac{19±\sqrt{393}}{16} when ± is plus. Add 19 to \sqrt{393}.

$$x=\frac{\sqrt{393}+19}{16}$$

Now solve the equation x=\frac{19±\sqrt{393}}{16} when ± is minus. Subtract \sqrt{393} from 19.

$$x=\frac{19-\sqrt{393}}{16}$$

The equation is now solved.

$$x=\frac{\sqrt{393}+19}{16} x=\frac{19-\sqrt{393}}{16}$$

Steps for Completing the Square

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

$$9x\left(x-2\right)=x^{2}+x+1$$

Use the distributive property to multiply 9x by x-2.

$$9x^{2}-18x=x^{2}+x+1$$

Subtract x^{2} from both sides.

$$9x^{2}-18x-x^{2}=x+1$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}-18x=x+1$$

Subtract x from both sides.

$$8x^{2}-18x-x=1$$

Combine -18x and -x to get -19x.

$$8x^{2}-19x=1$$

Divide both sides by 8.

$$\frac{8x^{2}-19x}{8}=\frac{1}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}-\frac{19}{8}x=\frac{1}{8}$$

Divide -\frac{19}{8}, the coefficient of the x term, by 2 to get -\frac{19}{16}. Then add the square of -\frac{19}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{19}{8}x+\left(-\frac{19}{16}\right)^{2}=\frac{1}{8}+\left(-\frac{19}{16}\right)^{2}$$

Square -\frac{19}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{19}{8}x+\frac{361}{256}=\frac{1}{8}+\frac{361}{256}$$

Add \frac{1}{8} to \frac{361}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{19}{8}x+\frac{361}{256}=\frac{393}{256}$$

Factor x^{2}-\frac{19}{8}x+\frac{361}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{19}{16}\right)^{2}=\frac{393}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{19}{16}\right)^{2}}=\sqrt{\frac{393}{256}}$$

Simplify.

$$x-\frac{19}{16}=\frac{\sqrt{393}}{16} x-\frac{19}{16}=-\frac{\sqrt{393}}{16}$$

Add \frac{19}{16} to both sides of the equation.

$$x=\frac{\sqrt{393}+19}{16} x=\frac{19-\sqrt{393}}{16}$$