Solve for p square root of (p+2)^2=-5

$\sqrt{{(p + 2)}^{2}} = - 5$

Step 1

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

${\sqrt{{(p + 2)}^{2}}}^{2} = {(- 5)}^{2}$

Step 2

Simplify each side of the equation.

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

${({(p + 2)}^{\frac{2}{2}})}^{2} = {(- 5)}^{2}$

Divide $2$ by $2$ .

${({(p + 2)}^{1})}^{2} = {(- 5)}^{2}$

Simplify the left side.

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

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

${(p + 2)}^{1 \cdot 2} = {(- 5)}^{2}$

Multiply $2$ by $1$ .

${(p + 2)}^{2} = {(- 5)}^{2}$

Simplify the right side.

Raise $- 5$ to the power of $2$ .

${(p + 2)}^{2} = 25$

Step 3

Solve for $p$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$p + 2 = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$p + 2 = \pm \sqrt{5^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$p + 2 = \pm 5$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$p + 2 = 5$

Move all terms not containing $p$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$p = 5 - 2$

Subtract $2$ from $5$ .

$p = 3$

Next, use the negative value of the $\pm$ to find the second solution.

$p + 2 = - 5$

Move all terms not containing $p$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$p = - 5 - 2$

Subtract $2$ from $- 5$ .

$p = - 7$

The complete solution is the result of both the positive and negative portions of the solution.

$p = 3, - 7$

Step 4

Exclude the solutions that do not make $\sqrt{{(p + 2)}^{2}} = - 5$ true.

$No solution$

Solve for p square root of (p+2)^2=-5

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