Solve 3/x^2-9+11/2x+6=1

$$\frac{ 3 }{ { x }^{ 2 } -9 } + \frac{ 11 }{ 2x+6 } = 1$$

Solve for x

$$x=1$$

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right)\left(x+3\right), the least common multiple of x^{2}-9,2x+6.

$$2\times 3+\left(x-3\right)\times 11=2\left(x-3\right)\left(x+3\right)$$

Multiply 2 and 3 to get 6.

$$6+\left(x-3\right)\times 11=2\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply x-3 by 11.

$$6+11x-33=2\left(x-3\right)\left(x+3\right)$$

Subtract 33 from 6 to get -27.

$$-27+11x=2\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply 2 by x-3.

$$-27+11x=\left(2x-6\right)\left(x+3\right)$$

Use the distributive property to multiply 2x-6 by x+3 and combine like terms.

$$-27+11x=2x^{2}-18$$

Subtract 2x^{2} from both sides.

$$-27+11x-2x^{2}=-18$$

Add 18 to both sides.

$$-27+11x-2x^{2}+18=0$$

Add -27 and 18 to get -9.

$$-9+11x-2x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}+11x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=11 ab=-2\left(-9\right)=18$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 11.

$$a=9 b=2$$

Rewrite -2x^{2}+11x-9 as \left(-2x^{2}+9x\right)+\left(2x-9\right).

$$\left(-2x^{2}+9x\right)+\left(2x-9\right)$$

Factor out -x in -2x^{2}+9x.

$$-x\left(2x-9\right)+2x-9$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(-x+1\right)$$

To find equation solutions, solve 2x-9=0 and -x+1=0.

$$x=\frac{9}{2} x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right)\left(x+3\right), the least common multiple of x^{2}-9,2x+6.

$$2\times 3+\left(x-3\right)\times 11=2\left(x-3\right)\left(x+3\right)$$

Multiply 2 and 3 to get 6.

$$6+\left(x-3\right)\times 11=2\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply x-3 by 11.

$$6+11x-33=2\left(x-3\right)\left(x+3\right)$$

Subtract 33 from 6 to get -27.

$$-27+11x=2\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply 2 by x-3.

$$-27+11x=\left(2x-6\right)\left(x+3\right)$$

Use the distributive property to multiply 2x-6 by x+3 and combine like terms.

$$-27+11x=2x^{2}-18$$

Subtract 2x^{2} from both sides.

$$-27+11x-2x^{2}=-18$$

Add 18 to both sides.

$$-27+11x-2x^{2}+18=0$$

Add -27 and 18 to get -9.

$$-9+11x-2x^{2}=0$$

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.

$$-2x^{2}+11x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 11 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-2\right)\left(-9\right)}}{2\left(-2\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-2\right)\left(-9\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-11±\sqrt{121+8\left(-9\right)}}{2\left(-2\right)}$$

Multiply 8 times -9.

$$x=\frac{-11±\sqrt{121-72}}{2\left(-2\right)}$$

Add 121 to -72.

$$x=\frac{-11±\sqrt{49}}{2\left(-2\right)}$$

Take the square root of 49.

$$x=\frac{-11±7}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-11±7}{-4}$$

Now solve the equation x=\frac{-11±7}{-4} when ± is plus. Add -11 to 7.

$$x=-\frac{4}{-4}$$

Divide -4 by -4.

$$x=1$$

Now solve the equation x=\frac{-11±7}{-4} when ± is minus. Subtract 7 from -11.

$$x=-\frac{18}{-4}$$

Reduce the fraction \frac{-18}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{2}$$

The equation is now solved.

$$x=1 x=\frac{9}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right)\left(x+3\right), the least common multiple of x^{2}-9,2x+6.

$$2\times 3+\left(x-3\right)\times 11=2\left(x-3\right)\left(x+3\right)$$

Multiply 2 and 3 to get 6.

$$6+\left(x-3\right)\times 11=2\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply x-3 by 11.

$$6+11x-33=2\left(x-3\right)\left(x+3\right)$$

Subtract 33 from 6 to get -27.

$$-27+11x=2\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply 2 by x-3.

$$-27+11x=\left(2x-6\right)\left(x+3\right)$$

Use the distributive property to multiply 2x-6 by x+3 and combine like terms.

$$-27+11x=2x^{2}-18$$

Subtract 2x^{2} from both sides.

$$-27+11x-2x^{2}=-18$$

Add 27 to both sides.

$$11x-2x^{2}=-18+27$$

Add -18 and 27 to get 9.

$$11x-2x^{2}=9$$

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.

$$-2x^{2}+11x=9$$

Divide both sides by -2.

$$\frac{-2x^{2}+11x}{-2}=\frac{9}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{11}{-2}x=\frac{9}{-2}$$

Divide 11 by -2.

$$x^{2}-\frac{11}{2}x=\frac{9}{-2}$$

Divide 9 by -2.

$$x^{2}-\frac{11}{2}x=-\frac{9}{2}$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-\frac{9}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-\frac{9}{2}+\frac{121}{16}$$

Add -\frac{9}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{49}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{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}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{7}{4} x-\frac{11}{4}=-\frac{7}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=\frac{9}{2} x=1$$