Precalculus Examples

$A = 2 p r (r + h)$

Step 1

Rewrite the equation as $2 p r (r + h) = A$ .

$2 p r (r + h) = A$

Step 2

Simplify $2 p r (r + h)$ .

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Step 2.1

Apply the distributive property.

$2 p r \cdot r + 2 p r h = A$

Step 2.2

Multiply $r$ by $r$ by adding the exponents.

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Step 2.2.1

Move $r$ .

$2 p (r \cdot r) + 2 p r h = A$

Step 2.2.2

Multiply $r$ by $r$ .

$2 p r^{2} + 2 p r h = A$

Step 3

Subtract $A$ from both sides of the equation.

$2 p r^{2} + 2 p r h - A = 0$

Step 4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5

Substitute the values $a = 2 p$ , $b = 2 p h$ , and $c = - A$ into the quadratic formula and solve for $r$ .

$\frac{- 2 p h \pm \sqrt{{(2 p h)}^{2} - 4 \cdot (2 p \cdot (- A))}}{2 (2 p)}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Add parentheses.

$r = \frac{- 2 p h \pm \sqrt{{(2 (p h))}^{2} - 4 \cdot (2 (p \cdot (- A)))}}{2 \cdot (2 p)}$

Step 6.1.2

Let $u = 2 (p \cdot (- A))$ . Substitute $u$ for all occurrences of $2 (p \cdot (- A))$ .

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Step 6.1.2.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 6.1.2.1.1

Apply the product rule to $2 p h$ .

$r = \frac{- 2 p h \pm \sqrt{{(2 p)}^{2} h^{2} - 4 \cdot u}}{2 \cdot (2 p)}$

Step 6.1.2.1.2

Apply the product rule to $2 p$ .

$r = \frac{- 2 p h \pm \sqrt{2^{2} p^{2} h^{2} - 4 \cdot u}}{2 \cdot (2 p)}$

Step 6.1.2.2

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

$r = \frac{- 2 p h \pm \sqrt{4 p^{2} h^{2} - 4 u}}{2 \cdot (2 p)}$

Step 6.1.3

Factor $4$ out of $4 p^{2} h^{2} - 4 u$ .

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Step 6.1.3.1

Factor $4$ out of $4 p^{2} h^{2}$ .

$r = \frac{- 2 p h \pm \sqrt{4 (p^{2} h^{2}) - 4 u}}{2 \cdot (2 p)}$

Step 6.1.3.2

Factor $4$ out of $- 4 u$ .

$r = \frac{- 2 p h \pm \sqrt{4 (p^{2} h^{2}) + 4 (- u)}}{2 \cdot (2 p)}$

Step 6.1.3.3

Factor $4$ out of $4 (p^{2} h^{2}) + 4 (- u)$ .

$r = \frac{- 2 p h \pm \sqrt{4 (p^{2} h^{2} - u)}}{2 \cdot (2 p)}$

Step 6.1.4

Replace all occurrences of $u$ with $2 (p \cdot (- A))$ .

$r = \frac{- 2 p h \pm \sqrt{4 (p^{2} h^{2} - (2 (p \cdot (- A))))}}{2 \cdot (2 p)}$

Step 6.1.5

Simplify each term.

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Step 6.1.5.1

Rewrite using the commutative property of multiplication.

$r = \frac{- 2 p h \pm \sqrt{4 (p^{2} h^{2} - (2 (- p A)))}}{2 \cdot (2 p)}$

Step 6.1.5.2

Multiply $- 1$ by $2$ .

$r = \frac{- 2 p h \pm \sqrt{4 (p^{2} h^{2} - (- 2 (p A)))}}{2 \cdot (2 p)}$

Step 6.1.5.3

Multiply $- 2$ by $- 1$ .

$r = \frac{- 2 p h \pm \sqrt{4 (p^{2} h^{2} + 2 p A)}}{2 \cdot (2 p)}$

Step 6.1.6

Factor $p$ out of $p^{2} h^{2} + 2 p A$ .

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Step 6.1.6.1

Factor $p$ out of $p^{2} h^{2}$ .

$r = \frac{- 2 p h \pm \sqrt{4 (p (p h^{2}) + 2 p A)}}{2 \cdot (2 p)}$

Step 6.1.6.2

Factor $p$ out of $2 p A$ .

$r = \frac{- 2 p h \pm \sqrt{4 (p (p h^{2}) + p (2 A))}}{2 \cdot (2 p)}$

Step 6.1.6.3

Factor $p$ out of $p (p h^{2}) + p (2 A)$ .

$r = \frac{- 2 p h \pm \sqrt{4 (p (p h^{2} + 2 A))}}{2 \cdot (2 p)}$

$r = \frac{- 2 p h \pm \sqrt{4 p (p h^{2} + 2 A)}}{2 \cdot (2 p)}$

Step 6.1.7

Rewrite $4 p (p h^{2} + 2 A)$ as $2^{2} (p (p h^{2} + 2 A))$ .

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Step 6.1.7.1

Rewrite $4$ as $2^{2}$ .

$r = \frac{- 2 p h \pm \sqrt{2^{2} p (p h^{2} + 2 A)}}{2 \cdot (2 p)}$

Step 6.1.7.2

Add parentheses.

$r = \frac{- 2 p h \pm \sqrt{2^{2} (p (p h^{2} + 2 A))}}{2 \cdot (2 p)}$

Step 6.1.8

Pull terms out from under the radical.

$r = \frac{- 2 p h \pm 2 \sqrt{p (p h^{2} + 2 A)}}{2 \cdot (2 p)}$

Step 6.2

Multiply $2$ by $2$ .

$r = \frac{- 2 p h \pm 2 \sqrt{p (p h^{2} + 2 A)}}{4 p}$

Step 6.3

Simplify $\frac{- 2 p h \pm 2 \sqrt{p (p h^{2} + 2 A)}}{4 p}$ .

$r = \frac{- p h \pm \sqrt{p (p h^{2} + 2 A)}}{2 p}$

Step 7

The final answer is the combination of both solutions.

$r = - \frac{p h - \sqrt{p (p h^{2} + 2 A)}}{2 p}$

$r = - \frac{p h + \sqrt{p (p h^{2} + 2 A)}}{2 p}$