Solve 2r/r-3=/r^2-9

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

Solve for r

$$r=\frac{\sqrt{11}-3}{2}\approx 0.158312395$$

$$r=\frac{-\sqrt{11}-3}{2}\approx -3.158312395$$

Steps Using the Quadratic Formula

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

$$\left(r+3\right)\times 2r=1$$

Use the distributive property to multiply r+3 by 2.

$$\left(2r+6\right)r=1$$

Use the distributive property to multiply 2r+6 by r.

$$2r^{2}+6r=1$$

Subtract 1 from both sides.

$$2r^{2}+6r-1=0$$

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

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

Square 6.

$$r=\frac{-6±\sqrt{36-4\times 2\left(-1\right)}}{2\times 2}$$

Multiply -4 times 2.

$$r=\frac{-6±\sqrt{36-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$r=\frac{-6±\sqrt{36+8}}{2\times 2}$$

Add 36 to 8.

$$r=\frac{-6±\sqrt{44}}{2\times 2}$$

Take the square root of 44.

$$r=\frac{-6±2\sqrt{11}}{2\times 2}$$

Multiply 2 times 2.

$$r=\frac{-6±2\sqrt{11}}{4}$$

Now solve the equation r=\frac{-6±2\sqrt{11}}{4} when ± is plus. Add -6 to 2\sqrt{11}.

$$r=\frac{2\sqrt{11}-6}{4}$$

Divide -6+2\sqrt{11} by 4.

$$r=\frac{\sqrt{11}-3}{2}$$

Now solve the equation r=\frac{-6±2\sqrt{11}}{4} when ± is minus. Subtract 2\sqrt{11} from -6.

$$r=\frac{-2\sqrt{11}-6}{4}$$

Divide -6-2\sqrt{11} by 4.

$$r=\frac{-\sqrt{11}-3}{2}$$

The equation is now solved.

$$r=\frac{\sqrt{11}-3}{2} r=\frac{-\sqrt{11}-3}{2}$$

Steps for Completing the Square

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

$$\left(r+3\right)\times 2r=1$$

Use the distributive property to multiply r+3 by 2.

$$\left(2r+6\right)r=1$$

Use the distributive property to multiply 2r+6 by r.

$$2r^{2}+6r=1$$

Divide both sides by 2.

$$\frac{2r^{2}+6r}{2}=\frac{1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$r^{2}+\frac{6}{2}r=\frac{1}{2}$$

Divide 6 by 2.

$$r^{2}+3r=\frac{1}{2}$$

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.

$$r^{2}+3r+\left(\frac{3}{2}\right)^{2}=\frac{1}{2}+\left(\frac{3}{2}\right)^{2}$$

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

$$r^{2}+3r+\frac{9}{4}=\frac{1}{2}+\frac{9}{4}$$

Add \frac{1}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}+3r+\frac{9}{4}=\frac{11}{4}$$

Factor r^{2}+3r+\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}.

$$\left(r+\frac{3}{2}\right)^{2}=\frac{11}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+\frac{3}{2}\right)^{2}}=\sqrt{\frac{11}{4}}$$

Simplify.

$$r+\frac{3}{2}=\frac{\sqrt{11}}{2} r+\frac{3}{2}=-\frac{\sqrt{11}}{2}$$

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

$$r=\frac{\sqrt{11}-3}{2} r=\frac{-\sqrt{11}-3}{2}$$