Solve 8/p-p/2/4=3/4

$$\frac { \frac { 8 } { p } - \frac { p } { 2 } } { 4 } = \frac { 3 } { 4 }$$

Solve for p

$$p=-8$$

$$p=2$$

Steps Using Factoring By Grouping

Multiply both sides by 4.

$$\frac{8}{p}-\frac{p}{2}=\frac{3}{4}\times 4$$

Variable p cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4p, the least common multiple of p,2,4.

$$4\times 8-2pp=\frac{3}{4}\times 4\times 4p$$

Multiply 4 and 8 to get 32.

$$32-2pp=\frac{3}{4}\times 4\times 4p$$

Multiply p and p to get p^{2}.

$$32-2p^{2}=\frac{3}{4}\times 4\times 4p$$

Cancel out 4 and 4.

$$32-2p^{2}=3\times 4p$$

Multiply 3 and 4 to get 12.

$$32-2p^{2}=12p$$

Subtract 12p from both sides.

$$32-2p^{2}-12p=0$$

Divide both sides by 2.

$$16-p^{2}-6p=0$$

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

$$-p^{2}-6p+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -p^{2}+ap+bp+16. To find a and b, set up a system to be solved.

$$a+b=-6 ab=-16=-16$$

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 -16.

$$1,-16 2,-8 4,-4$$

Calculate the sum for each pair.

$$1-16=-15 2-8=-6 4-4=0$$

The solution is the pair that gives sum -6.

$$a=2 b=-8$$

Rewrite -p^{2}-6p+16 as \left(-p^{2}+2p\right)+\left(-8p+16\right).

$$\left(-p^{2}+2p\right)+\left(-8p+16\right)$$

Factor out p in the first and 8 in the second group.

$$p\left(-p+2\right)+8\left(-p+2\right)$$

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

$$\left(-p+2\right)\left(p+8\right)$$

To find equation solutions, solve -p+2=0 and p+8=0.

$$p=2 p=-8$$

Steps Using the Quadratic Formula

Multiply both sides by 4.

$$\frac{8}{p}-\frac{p}{2}=\frac{3}{4}\times 4$$

Variable p cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4p, the least common multiple of p,2,4.

$$4\times 8-2pp=\frac{3}{4}\times 4\times 4p$$

Multiply 4 and 8 to get 32.

$$32-2pp=\frac{3}{4}\times 4\times 4p$$

Multiply p and p to get p^{2}.

$$32-2p^{2}=\frac{3}{4}\times 4\times 4p$$

Cancel out 4 and 4.

$$32-2p^{2}=3\times 4p$$

Multiply 3 and 4 to get 12.

$$32-2p^{2}=12p$$

Subtract 12p from both sides.

$$32-2p^{2}-12p=0$$

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.

$$-2p^{2}-12p+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -12 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-2\right)\times 32}}{2\left(-2\right)}$$

Square -12.

$$p=\frac{-\left(-12\right)±\sqrt{144-4\left(-2\right)\times 32}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$p=\frac{-\left(-12\right)±\sqrt{144+8\times 32}}{2\left(-2\right)}$$

Multiply 8 times 32.

$$p=\frac{-\left(-12\right)±\sqrt{144+256}}{2\left(-2\right)}$$

Add 144 to 256.

$$p=\frac{-\left(-12\right)±\sqrt{400}}{2\left(-2\right)}$$

Take the square root of 400.

$$p=\frac{-\left(-12\right)±20}{2\left(-2\right)}$$

The opposite of -12 is 12.

$$p=\frac{12±20}{2\left(-2\right)}$$

Multiply 2 times -2.

$$p=\frac{12±20}{-4}$$

Now solve the equation p=\frac{12±20}{-4} when ± is plus. Add 12 to 20.

$$p=\frac{32}{-4}$$

Divide 32 by -4.

$$p=-8$$

Now solve the equation p=\frac{12±20}{-4} when ± is minus. Subtract 20 from 12.

$$p=-\frac{8}{-4}$$

Divide -8 by -4.

$$p=2$$

The equation is now solved.

$$p=-8 p=2$$

Steps for Completing the Square

Multiply both sides by 4.

$$\frac{8}{p}-\frac{p}{2}=\frac{3}{4}\times 4$$

Variable p cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4p, the least common multiple of p,2,4.

$$4\times 8-2pp=\frac{3}{4}\times 4\times 4p$$

Multiply 4 and 8 to get 32.

$$32-2pp=\frac{3}{4}\times 4\times 4p$$

Multiply p and p to get p^{2}.

$$32-2p^{2}=\frac{3}{4}\times 4\times 4p$$

Cancel out 4 and 4.

$$32-2p^{2}=3\times 4p$$

Multiply 3 and 4 to get 12.

$$32-2p^{2}=12p$$

Subtract 12p from both sides.

$$32-2p^{2}-12p=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$-2p^{2}-12p=-32$$

Divide both sides by -2.

$$\frac{-2p^{2}-12p}{-2}=-\frac{32}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$p^{2}+\left(-\frac{12}{-2}\right)p=-\frac{32}{-2}$$

Divide -12 by -2.

$$p^{2}+6p=-\frac{32}{-2}$$

Divide -32 by -2.

$$p^{2}+6p=16$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+6p+3^{2}=16+3^{2}$$

Square 3.

$$p^{2}+6p+9=16+9$$

Add 16 to 9.

$$p^{2}+6p+9=25$$

Factor p^{2}+6p+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p+3\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+3\right)^{2}}=\sqrt{25}$$

Simplify.

$$p+3=5 p+3=-5$$

Subtract 3 from both sides of the equation.

$$p=2 p=-8$$