Solve p+24/p=11

$$p + \frac { 24 } { p } = 11$$

Solve for p

$$p=3$$

$$p=8$$

Steps Using Factoring

Variable p cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by p.

$$pp+24=11p$$

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

$$p^{2}+24=11p$$

Subtract 11p from both sides.

$$p^{2}+24-11p=0$$

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

$$p^{2}-11p+24=0$$

To solve the equation, factor p^{2}-11p+24 using formula p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right). To find a and b, set up a system to be solved.

$$a+b=-11 ab=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-8 b=-3$$

Rewrite factored expression \left(p+a\right)\left(p+b\right) using the obtained values.

$$\left(p-8\right)\left(p-3\right)$$

To find equation solutions, solve p-8=0 and p-3=0.

$$p=8 p=3$$

Steps Using Factoring By Grouping

Variable p cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by p.

$$pp+24=11p$$

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

$$p^{2}+24=11p$$

Subtract 11p from both sides.

$$p^{2}+24-11p=0$$

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

$$p^{2}-11p+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp+24. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 24=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-8 b=-3$$

Rewrite p^{2}-11p+24 as \left(p^{2}-8p\right)+\left(-3p+24\right).

$$\left(p^{2}-8p\right)+\left(-3p+24\right)$$

Factor out p in the first and -3 in the second group.

$$p\left(p-8\right)-3\left(p-8\right)$$

Factor out common term p-8 by using distributive property.

$$\left(p-8\right)\left(p-3\right)$$

To find equation solutions, solve p-8=0 and p-3=0.

$$p=8 p=3$$

Steps Using the Quadratic Formula

Variable p cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by p.

$$pp+24=11p$$

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

$$p^{2}+24=11p$$

Subtract 11p from both sides.

$$p^{2}+24-11p=0$$

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.

$$p^{2}-11p+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 24}}{2}$$

Square -11.

$$p=\frac{-\left(-11\right)±\sqrt{121-4\times 24}}{2}$$

Multiply -4 times 24.

$$p=\frac{-\left(-11\right)±\sqrt{121-96}}{2}$$

Add 121 to -96.

$$p=\frac{-\left(-11\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$p=\frac{-\left(-11\right)±5}{2}$$

The opposite of -11 is 11.

$$p=\frac{11±5}{2}$$

Now solve the equation p=\frac{11±5}{2} when ± is plus. Add 11 to 5.

$$p=\frac{16}{2}$$

Divide 16 by 2.

$$p=8$$

Now solve the equation p=\frac{11±5}{2} when ± is minus. Subtract 5 from 11.

$$p=\frac{6}{2}$$

Divide 6 by 2.

$$p=3$$

The equation is now solved.

$$p=8 p=3$$

Steps for Completing the Square

Variable p cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by p.

$$pp+24=11p$$

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

$$p^{2}+24=11p$$

Subtract 11p from both sides.

$$p^{2}+24-11p=0$$

Subtract 24 from both sides. Anything subtracted from zero gives its negation.

$$p^{2}-11p=-24$$

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

$$p^{2}-11p+\left(-\frac{11}{2}\right)^{2}=-24+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-11p+\frac{121}{4}=-24+\frac{121}{4}$$

Add -24 to \frac{121}{4}.

$$p^{2}-11p+\frac{121}{4}=\frac{25}{4}$$

Factor p^{2}-11p+\frac{121}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-\frac{11}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{11}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$p-\frac{11}{2}=\frac{5}{2} p-\frac{11}{2}=-\frac{5}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$p=8 p=3$$