Solve 4/(x^2-4)=1/(x-3)+1/(x+2)

$$\frac { 4 } { ( x ^ { 2 } - 4 ) } = \frac { 1 } { ( x - 3 ) } + \frac { 1 } { ( x + 2 ) }$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{31}i+9}{4}\approx 2.25-1.391941091i$$

$$x=\frac{9+\sqrt{31}i}{4}\approx 2.25+1.391941091i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,2,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x-2\right)\left(x+2\right), the least common multiple of x^{2}-4,x-3,x+2.

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

Use the distributive property to multiply x-3 by 4.

$$4x-12=x^{2}-4+\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply x-3 by x-2 and combine like terms.

$$4x-12=x^{2}-4+x^{2}-5x+6$$

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

$$4x-12=2x^{2}-4-5x+6$$

Add -4 and 6 to get 2.

$$4x-12=2x^{2}+2-5x$$

Subtract 2x^{2} from both sides.

$$4x-12-2x^{2}=2-5x$$

Subtract 2 from both sides.

$$4x-12-2x^{2}-2=-5x$$

Subtract 2 from -12 to get -14.

$$4x-14-2x^{2}=-5x$$

Add 5x to both sides.

$$4x-14-2x^{2}+5x=0$$

Combine 4x and 5x to get 9x.

$$9x-14-2x^{2}=0$$

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.

$$-2x^{2}+9x-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 9 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-2\right)\left(-14\right)}}{2\left(-2\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-2\right)\left(-14\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-9±\sqrt{81+8\left(-14\right)}}{2\left(-2\right)}$$

Multiply 8 times -14.

$$x=\frac{-9±\sqrt{81-112}}{2\left(-2\right)}$$

Add 81 to -112.

$$x=\frac{-9±\sqrt{-31}}{2\left(-2\right)}$$

Take the square root of -31.

$$x=\frac{-9±\sqrt{31}i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-9±\sqrt{31}i}{-4}$$

Now solve the equation x=\frac{-9±\sqrt{31}i}{-4} when ± is plus. Add -9 to i\sqrt{31}.

$$x=\frac{-9+\sqrt{31}i}{-4}$$

Divide -9+i\sqrt{31} by -4.

$$x=\frac{-\sqrt{31}i+9}{4}$$

Now solve the equation x=\frac{-9±\sqrt{31}i}{-4} when ± is minus. Subtract i\sqrt{31} from -9.

$$x=\frac{-\sqrt{31}i-9}{-4}$$

Divide -9-i\sqrt{31} by -4.

$$x=\frac{9+\sqrt{31}i}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{31}i+9}{4} x=\frac{9+\sqrt{31}i}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,2,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x-2\right)\left(x+2\right), the least common multiple of x^{2}-4,x-3,x+2.

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

Use the distributive property to multiply x-3 by 4.

$$4x-12=x^{2}-4+\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply x-3 by x-2 and combine like terms.

$$4x-12=x^{2}-4+x^{2}-5x+6$$

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

$$4x-12=2x^{2}-4-5x+6$$

Add -4 and 6 to get 2.

$$4x-12=2x^{2}+2-5x$$

Subtract 2x^{2} from both sides.

$$4x-12-2x^{2}=2-5x$$

Add 5x to both sides.

$$4x-12-2x^{2}+5x=2$$

Combine 4x and 5x to get 9x.

$$9x-12-2x^{2}=2$$

Add 12 to both sides.

$$9x-2x^{2}=2+12$$

Add 2 and 12 to get 14.

$$9x-2x^{2}=14$$

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.

$$-2x^{2}+9x=14$$

Divide both sides by -2.

$$\frac{-2x^{2}+9x}{-2}=\frac{14}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{9}{-2}x=\frac{14}{-2}$$

Divide 9 by -2.

$$x^{2}-\frac{9}{2}x=\frac{14}{-2}$$

Divide 14 by -2.

$$x^{2}-\frac{9}{2}x=-7$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=-7+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=-7+\frac{81}{16}$$

Add -7 to \frac{81}{16}.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=-\frac{31}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{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}.

$$\left(x-\frac{9}{4}\right)^{2}=-\frac{31}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{-\frac{31}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{\sqrt{31}i}{4} x-\frac{9}{4}=-\frac{\sqrt{31}i}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=\frac{9+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i+9}{4}$$