Solve for p p = square root of 2-p

$p = \sqrt{2 - p}$

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

$\sqrt{2 - p} = p$

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$2 - p = p^{2}$

Solve for $p$ .

Subtract $p^{2}$ from both sides of the equation.

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

Factor the left side of the equation.

Factor $- 1$ out of $2 - p - p^{2}$ .

Reorder the expression.

Move $2$ .

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

Reorder $- p$ and $- p^{2}$ .

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

Factor $- 1$ out of $- p^{2}$ .

$- (p^{2}) - p + 2 = 0$

Factor $- 1$ out of $- p$ .

$- (p^{2}) - (p) + 2 = 0$

Rewrite $2$ as $- 1 (- 2)$ .

$- (p^{2}) - (p) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (p^{2}) - (p)$ .

$- (p^{2} + p) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (p^{2} + p) - 1 (- 2)$ .

$- (p^{2} + p - 2) = 0$

Factor.

Factor $p^{2} + p - 2$ 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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$- ((p - 1) (p + 2)) = 0$

Remove unnecessary parentheses.

$- (p - 1) (p + 2) = 0$

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

$p - 1 = 0$

$p + 2 = 0$

Set $p - 1$ equal to $0$ and solve for $p$ .

Set $p - 1$ equal to $0$ .

$p - 1 = 0$

Add $1$ to both sides of the equation.

$p = 1$

Set $p + 2$ equal to $0$ and solve for $p$ .

Set $p + 2$ equal to $0$ .

$p + 2 = 0$

Subtract $2$ from both sides of the equation.

$p = - 2$

The final solution is all the values that make $- (p - 1) (p + 2) = 0$ true.

$p = 1, - 2$

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

$p = 1$

Solve for p p = square root of 2-p

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