Solve 1/n^2+11n+30=1/n+5-4

$$\frac { 1 } { n ^ { 2 } + 11 n + 30 } = \frac { 1 } { n + 5 } - 4$$

Solve for n

$$n = -\frac{23}{4} = -5\frac{3}{4} = -5.75$$

Steps Using Factoring By Grouping

Variable n cannot be equal to any of the values -6,-5 since division by zero is not defined. Multiply both sides of the equation by \left(n+5\right)\left(n+6\right), the least common multiple of n^{2}+11n+30,n+5.

$$1=n+6+\left(n+5\right)\left(n+6\right)\left(-4\right)$$

Use the distributive property to multiply n+5 by n+6 and combine like terms.

$$1=n+6+\left(n^{2}+11n+30\right)\left(-4\right)$$

Use the distributive property to multiply n^{2}+11n+30 by -4.

$$1=n+6-4n^{2}-44n-120$$

Combine n and -44n to get -43n.

$$1=-43n+6-4n^{2}-120$$

Subtract 120 from 6 to get -114.

$$1=-43n-114-4n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-43n-114-4n^{2}=1$$

Subtract 1 from both sides.

$$-43n-114-4n^{2}-1=0$$

Subtract 1 from -114 to get -115.

$$-43n-115-4n^{2}=0$$

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

$$-4n^{2}-43n-115=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4n^{2}+an+bn-115. To find a and b, set up a system to be solved.

$$a+b=-43 ab=-4\left(-115\right)=460$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 460.

$$-1,-460 -2,-230 -4,-115 -5,-92 -10,-46 -20,-23$$

Calculate the sum for each pair.

$$-1-460=-461 -2-230=-232 -4-115=-119 -5-92=-97 -10-46=-56 -20-23=-43$$

The solution is the pair that gives sum -43.

$$a=-20 b=-23$$

Rewrite -4n^{2}-43n-115 as \left(-4n^{2}-20n\right)+\left(-23n-115\right).

$$\left(-4n^{2}-20n\right)+\left(-23n-115\right)$$

Factor out 4n in the first and 23 in the second group.

$$4n\left(-n-5\right)+23\left(-n-5\right)$$

Factor out common term -n-5 by using distributive property.

$$\left(-n-5\right)\left(4n+23\right)$$

To find equation solutions, solve -n-5=0 and 4n+23=0.

$$n=-5 n=-\frac{23}{4}$$

Variable n cannot be equal to -5.

$$n=-\frac{23}{4}$$

Steps Using the Quadratic Formula

Variable n cannot be equal to any of the values -6,-5 since division by zero is not defined. Multiply both sides of the equation by \left(n+5\right)\left(n+6\right), the least common multiple of n^{2}+11n+30,n+5.

$$1=n+6+\left(n+5\right)\left(n+6\right)\left(-4\right)$$

Use the distributive property to multiply n+5 by n+6 and combine like terms.

$$1=n+6+\left(n^{2}+11n+30\right)\left(-4\right)$$

Use the distributive property to multiply n^{2}+11n+30 by -4.

$$1=n+6-4n^{2}-44n-120$$

Combine n and -44n to get -43n.

$$1=-43n+6-4n^{2}-120$$

Subtract 120 from 6 to get -114.

$$1=-43n-114-4n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-43n-114-4n^{2}=1$$

Subtract 1 from both sides.

$$-43n-114-4n^{2}-1=0$$

Subtract 1 from -114 to get -115.

$$-43n-115-4n^{2}=0$$

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.

$$-4n^{2}-43n-115=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -43 for b, and -115 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-43\right)±\sqrt{\left(-43\right)^{2}-4\left(-4\right)\left(-115\right)}}{2\left(-4\right)}$$

Square -43.

$$n=\frac{-\left(-43\right)±\sqrt{1849-4\left(-4\right)\left(-115\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$n=\frac{-\left(-43\right)±\sqrt{1849+16\left(-115\right)}}{2\left(-4\right)}$$

Multiply 16 times -115.

$$n=\frac{-\left(-43\right)±\sqrt{1849-1840}}{2\left(-4\right)}$$

Add 1849 to -1840.

$$n=\frac{-\left(-43\right)±\sqrt{9}}{2\left(-4\right)}$$

Take the square root of 9.

$$n=\frac{-\left(-43\right)±3}{2\left(-4\right)}$$

The opposite of -43 is 43.

$$n=\frac{43±3}{2\left(-4\right)}$$

Multiply 2 times -4.

$$n=\frac{43±3}{-8}$$

Now solve the equation n=\frac{43±3}{-8} when ± is plus. Add 43 to 3.

$$n=\frac{46}{-8}$$

Reduce the fraction \frac{46}{-8} to lowest terms by extracting and canceling out 2.

$$n=-\frac{23}{4}$$

Now solve the equation n=\frac{43±3}{-8} when ± is minus. Subtract 3 from 43.

$$n=\frac{40}{-8}$$

Divide 40 by -8.

$$n=-5$$

The equation is now solved.

$$n=-\frac{23}{4} n=-5$$

Variable n cannot be equal to -5.

$$n=-\frac{23}{4}$$

Steps for Completing the Square

Variable n cannot be equal to any of the values -6,-5 since division by zero is not defined. Multiply both sides of the equation by \left(n+5\right)\left(n+6\right), the least common multiple of n^{2}+11n+30,n+5.

$$1=n+6+\left(n+5\right)\left(n+6\right)\left(-4\right)$$

Use the distributive property to multiply n+5 by n+6 and combine like terms.

$$1=n+6+\left(n^{2}+11n+30\right)\left(-4\right)$$

Use the distributive property to multiply n^{2}+11n+30 by -4.

$$1=n+6-4n^{2}-44n-120$$

Combine n and -44n to get -43n.

$$1=-43n+6-4n^{2}-120$$

Subtract 120 from 6 to get -114.

$$1=-43n-114-4n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-43n-114-4n^{2}=1$$

Add 114 to both sides.

$$-43n-4n^{2}=1+114$$

Add 1 and 114 to get 115.

$$-43n-4n^{2}=115$$

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.

$$-4n^{2}-43n=115$$

Divide both sides by -4.

$$\frac{-4n^{2}-43n}{-4}=\frac{115}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$n^{2}+\left(-\frac{43}{-4}\right)n=\frac{115}{-4}$$

Divide -43 by -4.

$$n^{2}+\frac{43}{4}n=\frac{115}{-4}$$

Divide 115 by -4.

$$n^{2}+\frac{43}{4}n=-\frac{115}{4}$$

Divide \frac{43}{4}, the coefficient of the x term, by 2 to get \frac{43}{8}. Then add the square of \frac{43}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{43}{4}n+\left(\frac{43}{8}\right)^{2}=-\frac{115}{4}+\left(\frac{43}{8}\right)^{2}$$

Square \frac{43}{8} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{43}{4}n+\frac{1849}{64}=-\frac{115}{4}+\frac{1849}{64}$$

Add -\frac{115}{4} to \frac{1849}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+\frac{43}{4}n+\frac{1849}{64}=\frac{9}{64}$$

Factor n^{2}+\frac{43}{4}n+\frac{1849}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{43}{8}\right)^{2}=\frac{9}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{43}{8}\right)^{2}}=\sqrt{\frac{9}{64}}$$

Simplify.

$$n+\frac{43}{8}=\frac{3}{8} n+\frac{43}{8}=-\frac{3}{8}$$

Subtract \frac{43}{8} from both sides of the equation.

$$n=-5 n=-\frac{23}{4}$$

Variable n cannot be equal to -5.

$$n=-\frac{23}{4}$$