Solve x-1/x+2=9/1

$$x - 1 / x + 2 = 9 / 1$$

Solve for x

$$x = \frac{\sqrt{53} + 7}{2} \approx 7.140054945$$

$$x=\frac{7-\sqrt{53}}{2}\approx -0.140054945$$

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 x.

$$xx-1+x\times 2=x\times 9$$

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

$$x^{2}-1+x\times 2=x\times 9$$

Subtract x\times 9 from both sides.

$$x^{2}-1+x\times 2-x\times 9=0$$

Combine x\times 2 and -x\times 9 to get -7x.

$$x^{2}-1-7x=0$$

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.

$$x^{2}-7x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$x=\frac{-\left(-7\right)±\sqrt{49+4}}{2}$$

Add 49 to 4.

$$x=\frac{-\left(-7\right)±\sqrt{53}}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{53}}{2}$$

Now solve the equation x=\frac{7±\sqrt{53}}{2} when ± is plus. Add 7 to \sqrt{53}.

$$x=\frac{\sqrt{53}+7}{2}$$

Now solve the equation x=\frac{7±\sqrt{53}}{2} when ± is minus. Subtract \sqrt{53} from 7.

$$x=\frac{7-\sqrt{53}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{53}+7}{2} x=\frac{7-\sqrt{53}}{2}$$

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 x.

$$xx-1+x\times 2=x\times 9$$

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

$$x^{2}-1+x\times 2=x\times 9$$

Subtract x\times 9 from both sides.

$$x^{2}-1+x\times 2-x\times 9=0$$

Combine x\times 2 and -x\times 9 to get -7x.

$$x^{2}-1-7x=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$x^{2}-7x=1$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=1+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=1+\frac{49}{4}$$

Add 1 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{53}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{53}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{53}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{53}}{2} x-\frac{7}{2}=-\frac{\sqrt{53}}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=\frac{\sqrt{53}+7}{2} x=\frac{7-\sqrt{53}}{2}$$