Solve 3x-3/x+3=x-3/x

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

Solve for x (complex solution)

$$x=\frac{3+3\sqrt{7}i}{4}\approx 0.75+1.984313483i$$

$$x=\frac{-3\sqrt{7}i+3}{4}\approx 0.75-1.984313483i$$

Steps Using the Quadratic Formula

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

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

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

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

Consider \left(x+3\right)\left(x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3x^{2}-3x=x^{2}-9$$

Subtract x^{2} from both sides.

$$3x^{2}-3x-x^{2}=-9$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}-3x=-9$$

Add 9 to both sides.

$$2x^{2}-3x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\times 9}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\times 9}}{2\times 2}$$

Multiply -8 times 9.

$$x=\frac{-\left(-3\right)±\sqrt{9-72}}{2\times 2}$$

Add 9 to -72.

$$x=\frac{-\left(-3\right)±\sqrt{-63}}{2\times 2}$$

Take the square root of -63.

$$x=\frac{-\left(-3\right)±3\sqrt{7}i}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±3\sqrt{7}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±3\sqrt{7}i}{4}$$

Now solve the equation x=\frac{3±3\sqrt{7}i}{4} when ± is plus. Add 3 to 3i\sqrt{7}.

$$x=\frac{3+3\sqrt{7}i}{4}$$

Now solve the equation x=\frac{3±3\sqrt{7}i}{4} when ± is minus. Subtract 3i\sqrt{7} from 3.

$$x=\frac{-3\sqrt{7}i+3}{4}$$

The equation is now solved.

$$x=\frac{3+3\sqrt{7}i}{4} x=\frac{-3\sqrt{7}i+3}{4}$$

Steps for Completing the Square

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

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

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

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

Consider \left(x+3\right)\left(x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3x^{2}-3x=x^{2}-9$$

Subtract x^{2} from both sides.

$$3x^{2}-3x-x^{2}=-9$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}-3x=-9$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{9}{2}+\frac{9}{16}$$

Add -\frac{9}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{63}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=-\frac{63}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{-\frac{63}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{3\sqrt{7}i}{4} x-\frac{3}{4}=-\frac{3\sqrt{7}i}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{3+3\sqrt{7}i}{4} x=\frac{-3\sqrt{7}i+3}{4}$$