Solve q5/q+3-3/q-3=3q-1/q^2-9

$$q \frac { 5 } { q + 3 } - \frac { 3 } { q - 3 } = \frac { 3 q - 1 } { q ^ { 2 } - 9 }$$

Solve for q

$$q = \frac{\sqrt{601} + 21}{10} \approx 4.551530134$$

$$q=\frac{21-\sqrt{601}}{10}\approx -0.351530134$$

Steps Using the Quadratic Formula

Variable q 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(q-3\right)\left(q+3\right), the least common multiple of q+3,q-3,q^{2}-9.

$$q\left(q-3\right)\times 5-\left(q+3\right)\times 3=3q-1$$

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

$$\left(q^{2}-3q\right)\times 5-\left(q+3\right)\times 3=3q-1$$

Use the distributive property to multiply q^{2}-3q by 5.

$$5q^{2}-15q-\left(q+3\right)\times 3=3q-1$$

Use the distributive property to multiply q+3 by 3.

$$5q^{2}-15q-\left(3q+9\right)=3q-1$$

To find the opposite of 3q+9, find the opposite of each term.

$$5q^{2}-15q-3q-9=3q-1$$

Combine -15q and -3q to get -18q.

$$5q^{2}-18q-9=3q-1$$

Subtract 3q from both sides.

$$5q^{2}-18q-9-3q=-1$$

Combine -18q and -3q to get -21q.

$$5q^{2}-21q-9=-1$$

Add 1 to both sides.

$$5q^{2}-21q-9+1=0$$

Add -9 and 1 to get -8.

$$5q^{2}-21q-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -21 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 5\left(-8\right)}}{2\times 5}$$

Square -21.

$$q=\frac{-\left(-21\right)±\sqrt{441-4\times 5\left(-8\right)}}{2\times 5}$$

Multiply -4 times 5.

$$q=\frac{-\left(-21\right)±\sqrt{441-20\left(-8\right)}}{2\times 5}$$

Multiply -20 times -8.

$$q=\frac{-\left(-21\right)±\sqrt{441+160}}{2\times 5}$$

Add 441 to 160.

$$q=\frac{-\left(-21\right)±\sqrt{601}}{2\times 5}$$

The opposite of -21 is 21.

$$q=\frac{21±\sqrt{601}}{2\times 5}$$

Multiply 2 times 5.

$$q=\frac{21±\sqrt{601}}{10}$$

Now solve the equation q=\frac{21±\sqrt{601}}{10} when ± is plus. Add 21 to \sqrt{601}.

$$q=\frac{\sqrt{601}+21}{10}$$

Now solve the equation q=\frac{21±\sqrt{601}}{10} when ± is minus. Subtract \sqrt{601} from 21.

$$q=\frac{21-\sqrt{601}}{10}$$

The equation is now solved.

$$q=\frac{\sqrt{601}+21}{10} q=\frac{21-\sqrt{601}}{10}$$

Steps for Completing the Square

Variable q 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(q-3\right)\left(q+3\right), the least common multiple of q+3,q-3,q^{2}-9.

$$q\left(q-3\right)\times 5-\left(q+3\right)\times 3=3q-1$$

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

$$\left(q^{2}-3q\right)\times 5-\left(q+3\right)\times 3=3q-1$$

Use the distributive property to multiply q^{2}-3q by 5.

$$5q^{2}-15q-\left(q+3\right)\times 3=3q-1$$

Use the distributive property to multiply q+3 by 3.

$$5q^{2}-15q-\left(3q+9\right)=3q-1$$

To find the opposite of 3q+9, find the opposite of each term.

$$5q^{2}-15q-3q-9=3q-1$$

Combine -15q and -3q to get -18q.

$$5q^{2}-18q-9=3q-1$$

Subtract 3q from both sides.

$$5q^{2}-18q-9-3q=-1$$

Combine -18q and -3q to get -21q.

$$5q^{2}-21q-9=-1$$

Add 9 to both sides.

$$5q^{2}-21q=-1+9$$

Add -1 and 9 to get 8.

$$5q^{2}-21q=8$$

Divide both sides by 5.

$$\frac{5q^{2}-21q}{5}=\frac{8}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$q^{2}-\frac{21}{5}q=\frac{8}{5}$$

Divide -\frac{21}{5}, the coefficient of the x term, by 2 to get -\frac{21}{10}. Then add the square of -\frac{21}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}-\frac{21}{5}q+\left(-\frac{21}{10}\right)^{2}=\frac{8}{5}+\left(-\frac{21}{10}\right)^{2}$$

Square -\frac{21}{10} by squaring both the numerator and the denominator of the fraction.

$$q^{2}-\frac{21}{5}q+\frac{441}{100}=\frac{8}{5}+\frac{441}{100}$$

Add \frac{8}{5} to \frac{441}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$q^{2}-\frac{21}{5}q+\frac{441}{100}=\frac{601}{100}$$

Factor q^{2}-\frac{21}{5}q+\frac{441}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q-\frac{21}{10}\right)^{2}=\frac{601}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q-\frac{21}{10}\right)^{2}}=\sqrt{\frac{601}{100}}$$

Simplify.

$$q-\frac{21}{10}=\frac{\sqrt{601}}{10} q-\frac{21}{10}=-\frac{\sqrt{601}}{10}$$

Add \frac{21}{10} to both sides of the equation.

$$q=\frac{\sqrt{601}+21}{10} q=\frac{21-\sqrt{601}}{10}$$