Solve 20p+8/p^2-16=3p/p-4-p-5/p+4

$$\frac { 20 p + 8 } { p ^ { 2 } - 16 } = \frac { 3 p } { p - 4 } - \frac { p - 5 } { p + 4 }$$

Solve for p

$$p = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Steps Using Factoring By Grouping

Variable p cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(p-4\right)\left(p+4\right), the least common multiple of p^{2}-16,p-4,p+4.

$$20p+8=\left(p+4\right)\times 3p-\left(p-4\right)\left(p-5\right)$$

Use the distributive property to multiply p+4 by 3.

$$20p+8=\left(3p+12\right)p-\left(p-4\right)\left(p-5\right)$$

Use the distributive property to multiply 3p+12 by p.

$$20p+8=3p^{2}+12p-\left(p-4\right)\left(p-5\right)$$

Use the distributive property to multiply p-4 by p-5 and combine like terms.

$$20p+8=3p^{2}+12p-\left(p^{2}-9p+20\right)$$

To find the opposite of p^{2}-9p+20, find the opposite of each term.

$$20p+8=3p^{2}+12p-p^{2}+9p-20$$

Combine 3p^{2} and -p^{2} to get 2p^{2}.

$$20p+8=2p^{2}+12p+9p-20$$

Combine 12p and 9p to get 21p.

$$20p+8=2p^{2}+21p-20$$

Subtract 2p^{2} from both sides.

$$20p+8-2p^{2}=21p-20$$

Subtract 21p from both sides.

$$20p+8-2p^{2}-21p=-20$$

Combine 20p and -21p to get -p.

$$-p+8-2p^{2}=-20$$

Add 20 to both sides.

$$-p+8-2p^{2}+20=0$$

Add 8 and 20 to get 28.

$$-p+28-2p^{2}=0$$

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

$$-2p^{2}-p+28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2p^{2}+ap+bp+28. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-2\times 28=-56$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -56.

$$1,-56 2,-28 4,-14 7,-8$$

Calculate the sum for each pair.

$$1-56=-55 2-28=-26 4-14=-10 7-8=-1$$

The solution is the pair that gives sum -1.

$$a=7 b=-8$$

Rewrite -2p^{2}-p+28 as \left(-2p^{2}+7p\right)+\left(-8p+28\right).

$$\left(-2p^{2}+7p\right)+\left(-8p+28\right)$$

Factor out -p in the first and -4 in the second group.

$$-p\left(2p-7\right)-4\left(2p-7\right)$$

Factor out common term 2p-7 by using distributive property.

$$\left(2p-7\right)\left(-p-4\right)$$

To find equation solutions, solve 2p-7=0 and -p-4=0.

$$p=\frac{7}{2} p=-4$$

Variable p cannot be equal to -4.

$$p=\frac{7}{2}$$

Steps Using the Quadratic Formula

Variable p cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(p-4\right)\left(p+4\right), the least common multiple of p^{2}-16,p-4,p+4.

$$20p+8=\left(p+4\right)\times 3p-\left(p-4\right)\left(p-5\right)$$

Use the distributive property to multiply p+4 by 3.

$$20p+8=\left(3p+12\right)p-\left(p-4\right)\left(p-5\right)$$

Use the distributive property to multiply 3p+12 by p.

$$20p+8=3p^{2}+12p-\left(p-4\right)\left(p-5\right)$$

Use the distributive property to multiply p-4 by p-5 and combine like terms.

$$20p+8=3p^{2}+12p-\left(p^{2}-9p+20\right)$$

To find the opposite of p^{2}-9p+20, find the opposite of each term.

$$20p+8=3p^{2}+12p-p^{2}+9p-20$$

Combine 3p^{2} and -p^{2} to get 2p^{2}.

$$20p+8=2p^{2}+12p+9p-20$$

Combine 12p and 9p to get 21p.

$$20p+8=2p^{2}+21p-20$$

Subtract 2p^{2} from both sides.

$$20p+8-2p^{2}=21p-20$$

Subtract 21p from both sides.

$$20p+8-2p^{2}-21p=-20$$

Combine 20p and -21p to get -p.

$$-p+8-2p^{2}=-20$$

Add 20 to both sides.

$$-p+8-2p^{2}+20=0$$

Add 8 and 20 to get 28.

$$-p+28-2p^{2}=0$$

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.

$$-2p^{2}-p+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -1 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)\times 28}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$p=\frac{-\left(-1\right)±\sqrt{1+8\times 28}}{2\left(-2\right)}$$

Multiply 8 times 28.

$$p=\frac{-\left(-1\right)±\sqrt{1+224}}{2\left(-2\right)}$$

Add 1 to 224.

$$p=\frac{-\left(-1\right)±\sqrt{225}}{2\left(-2\right)}$$

Take the square root of 225.

$$p=\frac{-\left(-1\right)±15}{2\left(-2\right)}$$

The opposite of -1 is 1.

$$p=\frac{1±15}{2\left(-2\right)}$$

Multiply 2 times -2.

$$p=\frac{1±15}{-4}$$

Now solve the equation p=\frac{1±15}{-4} when ± is plus. Add 1 to 15.

$$p=\frac{16}{-4}$$

Divide 16 by -4.

$$p=-4$$

Now solve the equation p=\frac{1±15}{-4} when ± is minus. Subtract 15 from 1.

$$p=-\frac{14}{-4}$$

Reduce the fraction \frac{-14}{-4} to lowest terms by extracting and canceling out 2.

$$p=\frac{7}{2}$$

The equation is now solved.

$$p=-4 p=\frac{7}{2}$$

Variable p cannot be equal to -4.

$$p=\frac{7}{2}$$

Steps for Completing the Square

Variable p cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(p-4\right)\left(p+4\right), the least common multiple of p^{2}-16,p-4,p+4.

$$20p+8=\left(p+4\right)\times 3p-\left(p-4\right)\left(p-5\right)$$

Use the distributive property to multiply p+4 by 3.

$$20p+8=\left(3p+12\right)p-\left(p-4\right)\left(p-5\right)$$

Use the distributive property to multiply 3p+12 by p.

$$20p+8=3p^{2}+12p-\left(p-4\right)\left(p-5\right)$$

Use the distributive property to multiply p-4 by p-5 and combine like terms.

$$20p+8=3p^{2}+12p-\left(p^{2}-9p+20\right)$$

To find the opposite of p^{2}-9p+20, find the opposite of each term.

$$20p+8=3p^{2}+12p-p^{2}+9p-20$$

Combine 3p^{2} and -p^{2} to get 2p^{2}.

$$20p+8=2p^{2}+12p+9p-20$$

Combine 12p and 9p to get 21p.

$$20p+8=2p^{2}+21p-20$$

Subtract 2p^{2} from both sides.

$$20p+8-2p^{2}=21p-20$$

Subtract 21p from both sides.

$$20p+8-2p^{2}-21p=-20$$

Combine 20p and -21p to get -p.

$$-p+8-2p^{2}=-20$$

Subtract 8 from both sides.

$$-p-2p^{2}=-20-8$$

Subtract 8 from -20 to get -28.

$$-p-2p^{2}=-28$$

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.

$$-2p^{2}-p=-28$$

Divide both sides by -2.

$$\frac{-2p^{2}-p}{-2}=-\frac{28}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$p^{2}+\left(-\frac{1}{-2}\right)p=-\frac{28}{-2}$$

Divide -1 by -2.

$$p^{2}+\frac{1}{2}p=-\frac{28}{-2}$$

Divide -28 by -2.

$$p^{2}+\frac{1}{2}p=14$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+\frac{1}{2}p+\left(\frac{1}{4}\right)^{2}=14+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$p^{2}+\frac{1}{2}p+\frac{1}{16}=14+\frac{1}{16}$$

Add 14 to \frac{1}{16}.

$$p^{2}+\frac{1}{2}p+\frac{1}{16}=\frac{225}{16}$$

Factor p^{2}+\frac{1}{2}p+\frac{1}{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}.

$$\left(p+\frac{1}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+\frac{1}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$p+\frac{1}{4}=\frac{15}{4} p+\frac{1}{4}=-\frac{15}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$p=\frac{7}{2} p=-4$$

Variable p cannot be equal to -4.

$$p=\frac{7}{2}$$