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

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

Solve for x

$$x = \frac{\sqrt{21} + 5}{2} \approx 4.791287847$$

$$x=\frac{5-\sqrt{21}}{2}\approx 0.208712153$$

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-3,x^{2}-4.

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

Use the distributive property to multiply x-3 by 2x+1 and combine like terms.

$$x^{2}-4=2x^{2}-5x-3$$

Subtract 2x^{2} from both sides.

$$x^{2}-4-2x^{2}=-5x-3$$

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

$$-x^{2}-4=-5x-3$$

Add 5x to both sides.

$$-x^{2}-4+5x=-3$$

Add 3 to both sides.

$$-x^{2}-4+5x+3=0$$

Add -4 and 3 to get -1.

$$-x^{2}-1+5x=0$$

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.

$$-x^{2}+5x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\left(-1\right)}}{2\left(-1\right)}$$

Multiply 4 times -1.

$$x=\frac{-5±\sqrt{25-4}}{2\left(-1\right)}$$

Add 25 to -4.

$$x=\frac{-5±\sqrt{21}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±\sqrt{21}}{-2}$$

Now solve the equation x=\frac{-5±\sqrt{21}}{-2} when ± is plus. Add -5 to \sqrt{21}.

$$x=\frac{\sqrt{21}-5}{-2}$$

Divide -5+\sqrt{21} by -2.

$$x=\frac{5-\sqrt{21}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{21}}{-2} when ± is minus. Subtract \sqrt{21} from -5.

$$x=\frac{-\sqrt{21}-5}{-2}$$

Divide -5-\sqrt{21} by -2.

$$x=\frac{\sqrt{21}+5}{2}$$

The equation is now solved.

$$x=\frac{5-\sqrt{21}}{2} x=\frac{\sqrt{21}+5}{2}$$

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-3,x^{2}-4.

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

Use the distributive property to multiply x-3 by 2x+1 and combine like terms.

$$x^{2}-4=2x^{2}-5x-3$$

Subtract 2x^{2} from both sides.

$$x^{2}-4-2x^{2}=-5x-3$$

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

$$-x^{2}-4=-5x-3$$

Add 5x to both sides.

$$-x^{2}-4+5x=-3$$

Add 4 to both sides.

$$-x^{2}+5x=-3+4$$

Add -3 and 4 to get 1.

$$-x^{2}+5x=1$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 5 by -1.

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

Divide 1 by -1.

$$x^{2}-5x=-1$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-1+\frac{25}{4}$$

Add -1 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{21}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{21}}{2} x-\frac{5}{2}=-\frac{\sqrt{21}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{21}+5}{2} x=\frac{5-\sqrt{21}}{2}$$