Solve 8-x=9-6/x

$$8 - x = 9 - \frac { 6 } { x }$$

Solve for x

$$x=-3$$

$$x=2$$

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.

$$x\times 8-xx=x\times 9-6$$

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

$$x\times 8-x^{2}=x\times 9-6$$

Subtract x\times 9 from both sides.

$$x\times 8-x^{2}-x\times 9=-6$$

Combine x\times 8 and -x\times 9 to get -x.

$$-x-x^{2}=-6$$

Add 6 to both sides.

$$-x-x^{2}+6=0$$

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.

$$-x^{2}-x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-1\right)}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-1\right)±5}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±5}{-2}$$

Now solve the equation x=\frac{1±5}{-2} when ± is plus. Add 1 to 5.

$$x=\frac{6}{-2}$$

Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{1±5}{-2} when ± is minus. Subtract 5 from 1.

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

The equation is now solved.

$$x=-3 x=2$$

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.

$$x\times 8-xx=x\times 9-6$$

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

$$x\times 8-x^{2}=x\times 9-6$$

Subtract x\times 9 from both sides.

$$x\times 8-x^{2}-x\times 9=-6$$

Combine x\times 8 and -x\times 9 to get -x.

$$-x-x^{2}=-6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}-x=-6$$

Divide both sides by -1.

$$\frac{-x^{2}-x}{-1}=-\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=-\frac{6}{-1}$$

Divide -1 by -1.

$$x^{2}+x=-\frac{6}{-1}$$

Divide -6 by -1.

$$x^{2}+x=6$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=6+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{25}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{5}{2} x+\frac{1}{2}=-\frac{5}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=2 x=-3$$