Precalculus Examples

$p = \frac{a^{2} R}{{(r + R)}^{2}}$

Step 1

Rewrite the equation as $\frac{a^{2} R}{{(r + R)}^{2}} = p$ .

$\frac{a^{2} R}{{(r + R)}^{2}} = p$

Step 2

Find the LCD of the terms in the equation.

Tap for more steps...

Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

${(r + R)}^{2}, 1$

Step 2.2

The LCM of one and any expression is the expression.

${(r + R)}^{2}$

Step 3

Multiply each term in $\frac{a^{2} R}{{(r + R)}^{2}} = p$ by ${(r + R)}^{2}$ to eliminate the fractions.

Tap for more steps...

Step 3.1

Multiply each term in $\frac{a^{2} R}{{(r + R)}^{2}} = p$ by ${(r + R)}^{2}$ .

$\frac{a^{2} R}{{(r + R)}^{2}} {(r + R)}^{2} = p {(r + R)}^{2}$

Step 3.2

Simplify the left side.

Tap for more steps...

Step 3.2.1

Cancel the common factor of ${(r + R)}^{2}$ .

Tap for more steps...

Step 3.2.1.1

Cancel the common factor.

$\frac{a^{2} R}{{(r + R)}^{2}} {(r + R)}^{2} = p {(r + R)}^{2}$

Step 3.2.1.2

Rewrite the expression.

$a^{2} R = p {(r + R)}^{2}$

Step 4

Solve the equation.

Tap for more steps...

Step 4.1

Simplify $p {(r + R)}^{2}$ .

Tap for more steps...

Step 4.1.1

Rewrite ${(r + R)}^{2}$ as $(r + R) (r + R)$ .

$a^{2} R = p ((r + R) (r + R))$

Step 4.1.2

Expand $(r + R) (r + R)$ using the FOIL Method.

Tap for more steps...

Step 4.1.2.1

Apply the distributive property.

$a^{2} R = p (r (r + R) + R (r + R))$

Step 4.1.2.2

Apply the distributive property.

$a^{2} R = p (r \cdot r + r R + R (r + R))$

Step 4.1.2.3

Apply the distributive property.

$a^{2} R = p (r \cdot r + r R + R r + R \cdot R)$

Step 4.1.3

Simplify and combine like terms.

Tap for more steps...

Step 4.1.3.1

Simplify each term.

Tap for more steps...

Step 4.1.3.1.1

Multiply $r$ by $r$ .

$a^{2} R = p (r^{2} + r R + R r + R \cdot R)$

Step 4.1.3.1.2

Multiply $R$ by $R$ .

$a^{2} R = p (r^{2} + r R + R r + R^{2})$

Step 4.1.3.2

Add $r R$ and $R r$ .

Tap for more steps...

Step 4.1.3.2.1

Reorder $R$ and $r$ .

$a^{2} R = p (r^{2} + r R + r R + R^{2})$

Step 4.1.3.2.2

Add $r R$ and $r R$ .

$a^{2} R = p (r^{2} + 2 r R + R^{2})$

Step 4.1.4

Apply the distributive property.

$a^{2} R = p r^{2} + p (2 r R) + p R^{2}$

Step 4.1.5

Rewrite using the commutative property of multiplication.

$a^{2} R = p r^{2} + 2 p r R + p R^{2}$

Step 4.2

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

$p r^{2} + 2 p r R + p R^{2} = a^{2} R$

Step 4.3

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

$p r^{2} + 2 p r R + p R^{2} - a^{2} R = 0$

Step 4.4

Use the quadratic formula to find the solutions.

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

Step 4.5

Substitute the values $a = p$ , $b = 2 p r - a^{2}$ , and $c = p r^{2}$ into the quadratic formula and solve for $R$ .

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

Step 4.6

Simplify the numerator.

Tap for more steps...

Step 4.6.1

Apply the distributive property.

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

Step 4.6.2

Multiply $2$ by $- 1$ .

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

Step 4.6.3

Multiply $- - a^{2}$ .

Tap for more steps...

Step 4.6.3.1

Multiply $- 1$ by $- 1$ .

$R = \frac{- 2 (p r) + 1 a^{2} \pm \sqrt{{(2 p r - a^{2})}^{2} - 4 \cdot p \cdot (p r^{2})}}{2 p}$

Step 4.6.3.2

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

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

Step 4.6.4

Rewrite $4 \cdot p \cdot (p r^{2})$ as ${(2 p r)}^{2}$ .

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

Step 4.6.5

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 p r - a^{2}$ and $b = 2 p r$ .

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

Step 4.6.6

Simplify.

Tap for more steps...

Step 4.6.6.1

Add $2 p r$ and $2 p r$ .

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

Step 4.6.6.2

Multiply $2$ by $- 1$ .

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

Step 4.6.6.3

Subtract $2 p r$ from $2 p r$ .

$R = \frac{- 2 p r + a^{2} \pm \sqrt{(- a^{2} + 4 p r) (- a^{2} + 0)}}{2 p}$

Step 4.6.6.4

Add $- a^{2}$ and $0$ .

$R = \frac{- 2 p r + a^{2} \pm \sqrt{(- a^{2} + 4 p r) \cdot (- 1 a^{2})}}{2 p}$

Step 4.6.6.5

Factor out negative.

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

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

Step 4.6.7

Rewrite $- (- a^{2} + 4 p r) a^{2}$ as $a^{2} \cdot (- (- a^{2} + 2^{2} p r))$ .

Tap for more steps...

Step 4.6.7.1

Move $- a^{2} + 4 p r$ .

$R = \frac{- 2 p r + a^{2} \pm \sqrt{- 1 a^{2} (- a^{2} + 4 p r)}}{2 p}$

Step 4.6.7.2

Reorder $- 1$ and $a^{2}$ .

$R = \frac{- 2 p r + a^{2} \pm \sqrt{a^{2} \cdot (- 1 (- a^{2} + 4 p r))}}{2 p}$

Step 4.6.7.3

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

$R = \frac{- 2 p r + a^{2} \pm \sqrt{a^{2} \cdot (- 1 (- a^{2} + 2^{2} p r))}}{2 p}$

Step 4.6.7.4

Add parentheses.

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

Step 4.6.8

Pull terms out from under the radical.

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

Step 4.6.9

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

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

Step 4.7

The final answer is the combination of both solutions.

$R = - \frac{2 p r - a^{2} - a \sqrt{- (- a^{2} + 4 p r)}}{2 p}$

$R = - \frac{2 p r - a^{2} + a \sqrt{- (- a^{2} + 4 p r)}}{2 p}$

$R = - \frac{2 p r - a^{2} - a \sqrt{- (- a^{2} + 4 p r)}}{2 p}$

$R = - \frac{2 p r - a^{2} + a \sqrt{- (- a^{2} + 4 p r)}}{2 p}$