Pre-Algebra Examples

$28 = 3 - {(1 + p)}^{2}$

Step 1

Rewrite the equation as $3 - {(1 + p)}^{2} = 28$ .

$3 - {(1 + p)}^{2} = 28$

Step 2

Subtract $3$ from both sides of the equation.

$- {(1 + p)}^{2} = 28 - 3$

Step 3

Subtract $3$ from $28$ .

$- {(1 + p)}^{2} = 25$

Step 4

Divide each term in $- {(1 + p)}^{2} = 25$ by $- 1$ and simplify.

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Step 4.1

Divide each term in $- {(1 + p)}^{2} = 25$ by $- 1$ .

$\frac{- {(1 + p)}^{2}}{- 1} = \frac{25}{- 1}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Dividing two negative values results in a positive value.

$\frac{{(1 + p)}^{2}}{1} = \frac{25}{- 1}$

Step 4.2.2

Divide ${(1 + p)}^{2}$ by $1$ .

${(1 + p)}^{2} = \frac{25}{- 1}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $25$ by $- 1$ .

${(1 + p)}^{2} = - 25$

Step 5

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

$1 + p = \pm \sqrt{- 25}$

Step 6

Simplify $\pm \sqrt{- 25}$ .

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Step 6.1

Rewrite $- 25$ as $- 1 (25)$ .

$1 + p = \pm \sqrt{- 1 (25)}$

Step 6.2

Rewrite $\sqrt{- 1 (25)}$ as $\sqrt{- 1} \cdot \sqrt{25}$ .

$1 + p = \pm \sqrt{- 1} \cdot \sqrt{25}$

Step 6.3

Rewrite $\sqrt{- 1}$ as $i$ .

$1 + p = \pm i \cdot \sqrt{25}$

Step 6.4

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

$1 + p = \pm i \cdot \sqrt{5^{2}}$

Step 6.5

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

$1 + p = \pm i \cdot 5$

Step 6.6

Move $5$ to the left of $i$ .

$1 + p = \pm 5 i$

Step 7

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

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Step 7.1

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

$1 + p = 5 i$

Step 7.2

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

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Step 7.2.1

Subtract $1$ from both sides of the equation.

$p = 5 i - 1$

Step 7.2.2

Reorder $5 i$ and $- 1$ .

$p = - 1 + 5 i$

Step 7.3

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

$1 + p = - 5 i$

Step 7.4

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

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Step 7.4.1

Subtract $1$ from both sides of the equation.

$p = - 5 i - 1$

Step 7.4.2

Reorder $- 5 i$ and $- 1$ .

$p = - 1 - 5 i$

Step 7.5

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

$p = - 1 + 5 i, - 1 - 5 i$