Solve for p p-2 square root of p=8

$p - 2 \sqrt{p} = 8$

Subtract $p$ from both sides of the equation.

$- 2 \sqrt{p} = 8 - p$

To remove the radical on the left side of the equation, square both sides of the equation.

${(- 2 \sqrt{p})}^{2} = {(8 - p)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{p}$ as $p^{\frac{1}{2}}$ .

${(- 2 p^{\frac{1}{2}})}^{2} = {(8 - p)}^{2}$

Simplify the left side.

Simplify ${(- 2 p^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- 2 p^{\frac{1}{2}}$ .

${(- 2)}^{2} {(p^{\frac{1}{2}})}^{2} = {(8 - p)}^{2}$

Raise $- 2$ to the power of $2$ .

$4 {(p^{\frac{1}{2}})}^{2} = {(8 - p)}^{2}$

Multiply the exponents in ${(p^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 p^{\frac{1}{2} \cdot 2} = {(8 - p)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 p^{\frac{1}{2} \cdot 2} = {(8 - p)}^{2}$

Rewrite the expression.

$4 p^{1} = {(8 - p)}^{2}$

Simplify.

$4 p = {(8 - p)}^{2}$

Simplify the right side.

Simplify ${(8 - p)}^{2}$ .

Rewrite ${(8 - p)}^{2}$ as $(8 - p) (8 - p)$ .

$4 p = (8 - p) (8 - p)$

Expand $(8 - p) (8 - p)$ using the FOIL Method.

Apply the distributive property.

$4 p = 8 (8 - p) - p (8 - p)$

Apply the distributive property.

$4 p = 8 \cdot 8 + 8 (- p) - p (8 - p)$

Apply the distributive property.

$4 p = 8 \cdot 8 + 8 (- p) - p \cdot 8 - p (- p)$

Simplify and combine like terms.

Simplify each term.

Multiply $8$ by $8$ .

$4 p = 64 + 8 (- p) - p \cdot 8 - p (- p)$

Multiply $- 1$ by $8$ .

$4 p = 64 - 8 p - p \cdot 8 - p (- p)$

Multiply $8$ by $- 1$ .

$4 p = 64 - 8 p - 8 p - p (- p)$

Rewrite using the commutative property of multiplication.

$4 p = 64 - 8 p - 8 p - 1 \cdot - 1 p \cdot p$

Multiply $p$ by $p$ by adding the exponents.

Move $p$ .

$4 p = 64 - 8 p - 8 p - 1 \cdot - 1 (p \cdot p)$

Multiply $p$ by $p$ .

$4 p = 64 - 8 p - 8 p - 1 \cdot - 1 p^{2}$

Multiply $- 1$ by $- 1$ .

$4 p = 64 - 8 p - 8 p + 1 p^{2}$

Multiply $p^{2}$ by $1$ .

$4 p = 64 - 8 p - 8 p + p^{2}$

Subtract $8 p$ from $- 8 p$ .

$4 p = 64 - 16 p + p^{2}$

Solve for $p$ .

Since $p$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$64 - 16 p + p^{2} = 4 p$

Move all terms containing $p$ to the left side of the equation.

Subtract $4 p$ from both sides of the equation.

$64 - 16 p + p^{2} - 4 p = 0$

Subtract $4 p$ from $- 16 p$ .

$64 + p^{2} - 20 p = 0$

Factor $64 + p^{2} - 20 p$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $64$ and whose sum is $- 20$ .

$- 16, - 4$

Write the factored form using these integers.

$(p - 16) (p - 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$p - 16 = 0$

$p - 4 = 0$

Set $p - 16$ equal to $0$ and solve for $p$ .

Set $p - 16$ equal to $0$ .

$p - 16 = 0$

Add $16$ to both sides of the equation.

$p = 16$

Set $p - 4$ equal to $0$ and solve for $p$ .

Set $p - 4$ equal to $0$ .

$p - 4 = 0$

Add $4$ to both sides of the equation.

$p = 4$

The final solution is all the values that make $(p - 16) (p - 4) = 0$ true.

$p = 16, 4$

Exclude the solutions that do not make $p - 2 \sqrt{p} = 8$ true.

$p = 16$

Solve for p p-2 square root of p=8

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