Solve 3/x^2-3x+1/x=2x/x^2-6x+9

$$\frac { 3 } { x ^ { 2 } - 3 x } + \frac { 1 } { x } = \frac { 2 x } { x ^ { 2 } - 6 x + 9 }$$

Solve for x

$$x=-3$$

Steps Using Factoring

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)^{2}, the least common multiple of x^{2}-3x,x,x^{2}-6x+9.

$$\left(x-3\right)\times 3+\left(x-3\right)^{2}=x\times 2x$$

Use the distributive property to multiply x-3 by 3.

$$3x-9+\left(x-3\right)^{2}=x\times 2x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$3x-9+x^{2}-6x+9=x\times 2x$$

Combine 3x and -6x to get -3x.

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

Add -9 and 9 to get 0.

$$-3x+x^{2}=x\times 2x$$

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

$$-3x+x^{2}=x^{2}\times 2$$

Subtract x^{2}\times 2 from both sides.

$$-3x+x^{2}-x^{2}\times 2=0$$

Combine x^{2} and -x^{2}\times 2 to get -x^{2}.

$$-3x-x^{2}=0$$

Factor out x.

$$x\left(-3-x\right)=0$$

To find equation solutions, solve x=0 and -3-x=0.

$$x=0 x=-3$$

Variable x cannot be equal to 0.

$$x=-3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)^{2}, the least common multiple of x^{2}-3x,x,x^{2}-6x+9.

$$\left(x-3\right)\times 3+\left(x-3\right)^{2}=x\times 2x$$

Use the distributive property to multiply x-3 by 3.

$$3x-9+\left(x-3\right)^{2}=x\times 2x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$3x-9+x^{2}-6x+9=x\times 2x$$

Combine 3x and -6x to get -3x.

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

Add -9 and 9 to get 0.

$$-3x+x^{2}=x\times 2x$$

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

$$-3x+x^{2}=x^{2}\times 2$$

Subtract x^{2}\times 2 from both sides.

$$-3x+x^{2}-x^{2}\times 2=0$$

Combine x^{2} and -x^{2}\times 2 to get -x^{2}.

$$-3x-x^{2}=0$$

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.

$$-x^{2}-3x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\left(-1\right)}$$

Take the square root of \left(-3\right)^{2}.

$$x=\frac{-\left(-3\right)±3}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±3}{-2}$$

Now solve the equation x=\frac{3±3}{-2} when ± is plus. Add 3 to 3.

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

Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{3±3}{-2} when ± is minus. Subtract 3 from 3.

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

The equation is now solved.

$$x=-3 x=0$$

Variable x cannot be equal to 0.

$$x=-3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)^{2}, the least common multiple of x^{2}-3x,x,x^{2}-6x+9.

$$\left(x-3\right)\times 3+\left(x-3\right)^{2}=x\times 2x$$

Use the distributive property to multiply x-3 by 3.

$$3x-9+\left(x-3\right)^{2}=x\times 2x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$3x-9+x^{2}-6x+9=x\times 2x$$

Combine 3x and -6x to get -3x.

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

Add -9 and 9 to get 0.

$$-3x+x^{2}=x\times 2x$$

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

$$-3x+x^{2}=x^{2}\times 2$$

Subtract x^{2}\times 2 from both sides.

$$-3x+x^{2}-x^{2}\times 2=0$$

Combine x^{2} and -x^{2}\times 2 to get -x^{2}.

$$-3x-x^{2}=0$$

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.

$$-x^{2}-3x=0$$

Divide both sides by -1.

$$\frac{-x^{2}-3x}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{3}{-1}\right)x=\frac{0}{-1}$$

Divide -3 by -1.

$$x^{2}+3x=\frac{0}{-1}$$

Divide 0 by -1.

$$x^{2}+3x=0$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{3}{2} x+\frac{3}{2}=-\frac{3}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=0 x=-3$$

Variable x cannot be equal to 0.

$$x=-3$$