Precalculus Examples

$\frac{{(y - A)}^{2}}{B^{2}} + \frac{{(x - C)}^{2}}{D^{2}} = 1$

Step 1

Subtract $\frac{{(x - C)}^{2}}{D^{2}}$ from both sides of the equation.

$\frac{{(y - A)}^{2}}{B^{2}} = 1 - \frac{{(x - C)}^{2}}{D^{2}}$

Step 2

Move all terms containing variables to the left side of the equation.

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Step 2.1

Add $\frac{{(x - C)}^{2}}{D^{2}}$ to both sides of the equation.

$\frac{{(y - A)}^{2}}{B^{2}} + \frac{{(x - C)}^{2}}{D^{2}} = 1$

Step 2.2

To write $\frac{{(y - A)}^{2}}{B^{2}}$ as a fraction with a common denominator, multiply by $\frac{D^{2}}{D^{2}}$ .

$\frac{{(y - A)}^{2}}{B^{2}} \cdot \frac{D^{2}}{D^{2}} + \frac{{(x - C)}^{2}}{D^{2}} = 1$

Step 2.3

To write $\frac{{(x - C)}^{2}}{D^{2}}$ as a fraction with a common denominator, multiply by $\frac{B^{2}}{B^{2}}$ .

$\frac{{(y - A)}^{2}}{B^{2}} \cdot \frac{D^{2}}{D^{2}} + \frac{{(x - C)}^{2}}{D^{2}} \cdot \frac{B^{2}}{B^{2}} = 1$

Step 2.4

Write each expression with a common denominator of $B^{2} D^{2}$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.4.1

Multiply $\frac{{(y - A)}^{2}}{B^{2}}$ by $\frac{D^{2}}{D^{2}}$ .

$\frac{{(y - A)}^{2} D^{2}}{B^{2} D^{2}} + \frac{{(x - C)}^{2}}{D^{2}} \cdot \frac{B^{2}}{B^{2}} = 1$

Step 2.4.2

Multiply $\frac{{(x - C)}^{2}}{D^{2}}$ by $\frac{B^{2}}{B^{2}}$ .

$\frac{{(y - A)}^{2} D^{2}}{B^{2} D^{2}} + \frac{{(x - C)}^{2} B^{2}}{D^{2} B^{2}} = 1$

Step 2.4.3

Reorder the factors of $D^{2} B^{2}$ .

$\frac{{(y - A)}^{2} D^{2}}{B^{2} D^{2}} + \frac{{(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.5

Combine the numerators over the common denominator.

$\frac{{(y - A)}^{2} D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6

Simplify the numerator.

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Step 2.6.1

Rewrite ${(y - A)}^{2}$ as $(y - A) (y - A)$ .

$\frac{(y - A) (y - A) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.2

Expand $(y - A) (y - A)$ using the FOIL Method.

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Step 2.6.2.1

Apply the distributive property.

$\frac{(y (y - A) - A (y - A)) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.2.2

Apply the distributive property.

$\frac{(y \cdot y + y (- A) - A (y - A)) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.2.3

Apply the distributive property.

$\frac{(y \cdot y + y (- A) - A y - A (- A)) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.3

Simplify and combine like terms.

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Step 2.6.3.1

Simplify each term.

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Step 2.6.3.1.1

Multiply $y$ by $y$ .

$\frac{(y^{2} + y (- A) - A y - A (- A)) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.3.1.2

Rewrite using the commutative property of multiplication.

$\frac{(y^{2} - y A - A y - A (- A)) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.3.1.3

Rewrite using the commutative property of multiplication.

$\frac{(y^{2} - y A - A y - 1 \cdot - 1 A \cdot A) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.3.1.4

Multiply $A$ by $A$ by adding the exponents.

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Step 2.6.3.1.4.1

Move $A$ .

$\frac{(y^{2} - y A - A y - 1 \cdot - 1 (A \cdot A)) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.3.1.4.2

Multiply $A$ by $A$ .

$\frac{(y^{2} - y A - A y - 1 \cdot - 1 A^{2}) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.3.1.5

Multiply $- 1$ by $- 1$ .

$\frac{(y^{2} - y A - A y + 1 A^{2}) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.3.1.6

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

$\frac{(y^{2} - y A - A y + A^{2}) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.3.2

Subtract $A y$ from $- y A$ .

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Step 2.6.3.2.1

Move $A$ .

$\frac{(y^{2} - y A - 1 y A + A^{2}) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.3.2.2

Subtract $y A$ from $- y A$ .

$\frac{(y^{2} - 2 y A + A^{2}) D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.4

Apply the distributive property.

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + {(x - C)}^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.5

Rewrite ${(x - C)}^{2}$ as $(x - C) (x - C)$ .

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x - C) (x - C) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.6

Expand $(x - C) (x - C)$ using the FOIL Method.

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Step 2.6.6.1

Apply the distributive property.

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x (x - C) - C (x - C)) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.6.2

Apply the distributive property.

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x \cdot x + x (- C) - C (x - C)) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.6.3

Apply the distributive property.

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x \cdot x + x (- C) - C x - C (- C)) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.7

Simplify and combine like terms.

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Step 2.6.7.1

Simplify each term.

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Step 2.6.7.1.1

Multiply $x$ by $x$ .

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x^{2} + x (- C) - C x - C (- C)) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.7.1.2

Rewrite using the commutative property of multiplication.

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x^{2} - x C - C x - C (- C)) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.7.1.3

Rewrite using the commutative property of multiplication.

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x^{2} - x C - C x - 1 \cdot - 1 C \cdot C) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.7.1.4

Multiply $C$ by $C$ by adding the exponents.

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Step 2.6.7.1.4.1

Move $C$ .

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x^{2} - x C - C x - 1 \cdot - 1 (C \cdot C)) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.7.1.4.2

Multiply $C$ by $C$ .

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x^{2} - x C - C x - 1 \cdot - 1 C^{2}) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.7.1.5

Multiply $- 1$ by $- 1$ .

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x^{2} - x C - C x + 1 C^{2}) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.7.1.6

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

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x^{2} - x C - C x + C^{2}) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.7.2

Subtract $C x$ from $- x C$ .

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Step 2.6.7.2.1

Move $x$ .

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x^{2} - C x - C x + C^{2}) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.7.2.2

Subtract $C x$ from $- C x$ .

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + (x^{2} - 2 C x + C^{2}) B^{2}}{B^{2} D^{2}} = 1$

Step 2.6.8

Apply the distributive property.

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + x^{2} B^{2} - 2 C x B^{2} + C^{2} B^{2}}{B^{2} D^{2}} = 1$

Step 3

Multiply both sides by $B^{2} D^{2}$ .

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + x^{2} B^{2} - 2 C x B^{2} + C^{2} B^{2}}{B^{2} D^{2}} (B^{2} D^{2}) = 1 (B^{2} D^{2})$

Step 4

Simplify.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Simplify $\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + x^{2} B^{2} - 2 C x B^{2} + C^{2} B^{2}}{B^{2} D^{2}} (B^{2} D^{2})$ .

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Step 4.1.1.1

Cancel the common factor of $B^{2} D^{2}$ .

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Step 4.1.1.1.1

Cancel the common factor.

$\frac{y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + x^{2} B^{2} - 2 C x B^{2} + C^{2} B^{2}}{B^{2} D^{2}} (B^{2} D^{2}) = 1 (B^{2} D^{2})$

Step 4.1.1.1.2

Rewrite the expression.

$y^{2} D^{2} - 2 y A D^{2} + A^{2} D^{2} + x^{2} B^{2} - 2 C x B^{2} + C^{2} B^{2} = 1 (B^{2} D^{2})$

Step 4.1.1.2

Simplify the expression.

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Step 4.1.1.2.1

Reorder $y^{2}$ and $D^{2}$ .

$D^{2} y^{2} - 2 y A D^{2} + A^{2} D^{2} + x^{2} B^{2} - 2 C x B^{2} + C^{2} B^{2} = 1 (B^{2} D^{2})$

Step 4.1.1.2.2

Move $A$ .

$D^{2} y^{2} - 2 y D^{2} A + A^{2} D^{2} + x^{2} B^{2} - 2 C x B^{2} + C^{2} B^{2} = 1 (B^{2} D^{2})$

Step 4.1.1.2.3

Move $y$ .

$D^{2} y^{2} - 2 D^{2} y A + A^{2} D^{2} + x^{2} B^{2} - 2 C x B^{2} + C^{2} B^{2} = 1 (B^{2} D^{2})$

Step 4.1.1.2.4

Reorder $A^{2}$ and $D^{2}$ .

$D^{2} y^{2} - 2 D^{2} y A + D^{2} A^{2} + x^{2} B^{2} - 2 C x B^{2} + C^{2} B^{2} = 1 (B^{2} D^{2})$

Step 4.1.1.2.5

Move $D^{2} y^{2}$ .

$- 2 D^{2} y A + D^{2} A^{2} + x^{2} B^{2} - 2 C x B^{2} + C^{2} B^{2} + D^{2} y^{2} = 1 (B^{2} D^{2})$

Step 4.1.1.2.6

Move $x^{2} B^{2}$ .

$- 2 D^{2} y A + D^{2} A^{2} - 2 C x B^{2} + C^{2} B^{2} + x^{2} B^{2} + D^{2} y^{2} = 1 (B^{2} D^{2})$

Step 4.1.1.2.7

Move $- 2 C x B^{2}$ .

$- 2 D^{2} y A + D^{2} A^{2} + C^{2} B^{2} - 2 C x B^{2} + x^{2} B^{2} + D^{2} y^{2} = 1 (B^{2} D^{2})$

Step 4.1.1.2.8

Reorder $- 2 D^{2} y A$ and $D^{2} A^{2}$ .

$D^{2} A^{2} - 2 D^{2} y A + C^{2} B^{2} - 2 C x B^{2} + x^{2} B^{2} + D^{2} y^{2} = 1 (B^{2} D^{2})$

Step 4.2

Simplify the right side.

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Step 4.2.1

Multiply $B^{2} D^{2}$ by $1$ .

$D^{2} A^{2} - 2 D^{2} y A + C^{2} B^{2} - 2 C x B^{2} + x^{2} B^{2} + D^{2} y^{2} = B^{2} D^{2}$

Step 5

Solve for $A$ .

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Step 5.1

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

$D^{2} A^{2} - 2 D^{2} y A + C^{2} B^{2} - 2 C x B^{2} + x^{2} B^{2} + D^{2} y^{2} - B^{2} D^{2} = 0$

Step 5.2

Use the quadratic formula to find the solutions.

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

Step 5.3

Substitute the values $a = D^{2}$ , $b = - 2 D^{2} y$ , and $c = C^{2} B^{2} - 2 C x B^{2} + x^{2} B^{2} + D^{2} y^{2} - B^{2} D^{2}$ into the quadratic formula and solve for $A$ .

$\frac{2 D^{2} y \pm \sqrt{{(- 2 D^{2} y)}^{2} - 4 \cdot (D^{2} \cdot (C^{2} B^{2} - 2 C x B^{2} + x^{2} B^{2} + D^{2} y^{2} - B^{2} D^{2}))}}{2 D^{2}}$

Step 5.4

Simplify.

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Step 5.4.1

Simplify the numerator.

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Step 5.4.1.1

Rewrite $B^{2} D^{2}$ as ${(B D)}^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{{(- 2 (D^{2} y))}^{2} - 4 \cdot (D^{2} \cdot ({(C B)}^{2} - 2 (C (x B^{2})) + {(x B)}^{2} + {(D y)}^{2} - {(B D)}^{2}))}}{2 D^{2}}$

Step 5.4.1.2

Let $u = D^{2} \cdot ({(C B)}^{2} - 2 (C (x B^{2})) + {(x B)}^{2} + {(D y)}^{2} - {(B D)}^{2})$ . Substitute $u$ for all occurrences of $D^{2} \cdot ({(C B)}^{2} - 2 (C (x B^{2})) + {(x B)}^{2} + {(D y)}^{2} - {(B D)}^{2})$ .

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Step 5.4.1.2.1

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

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Step 5.4.1.2.1.1

Apply the product rule to $- 2 D^{2} y$ .

$A = \frac{2 D^{2} y \pm \sqrt{{(- 2 D^{2})}^{2} y^{2} - 4 \cdot u}}{2 D^{2}}$

Step 5.4.1.2.1.2

Apply the product rule to $- 2 D^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{{(- 2)}^{2} {(D^{2})}^{2} y^{2} - 4 \cdot u}}{2 D^{2}}$

Step 5.4.1.2.2

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

$A = \frac{2 D^{2} y \pm \sqrt{4 {(D^{2})}^{2} y^{2} - 4 \cdot u}}{2 D^{2}}$

Step 5.4.1.2.3

Multiply the exponents in ${(D^{2})}^{2}$ .

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Step 5.4.1.2.3.1

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

$A = \frac{2 D^{2} y \pm \sqrt{4 D^{2 \cdot 2} y^{2} - 4 \cdot u}}{2 D^{2}}$

Step 5.4.1.2.3.2

Multiply $2$ by $2$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 D^{4} y^{2} - 4 \cdot u}}{2 D^{2}}$

$A = \frac{2 D^{2} y \pm \sqrt{4 D^{4} y^{2} - 4 u}}{2 D^{2}}$

Step 5.4.1.3

Factor $4$ out of $4 D^{4} y^{2} - 4 u$ .

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Step 5.4.1.3.1

Factor $4$ out of $4 D^{4} y^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2}) - 4 u}}{2 D^{2}}$

Step 5.4.1.3.2

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

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2}) + 4 (- u)}}{2 D^{2}}$

Step 5.4.1.3.3

Factor $4$ out of $4 (D^{4} y^{2}) + 4 (- u)$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - u)}}{2 D^{2}}$

Step 5.4.1.4

Replace all occurrences of $u$ with $D^{2} \cdot ({(C B)}^{2} - 2 (C (x B^{2})) + {(x B)}^{2} + {(D y)}^{2} - {(B D)}^{2})$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} \cdot ({(C B)}^{2} - 2 (C (x B^{2})) + {(x B)}^{2} + {(D y)}^{2} - {(B D)}^{2})))}}{2 D^{2}}$

Step 5.4.1.5

Simplify.

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Step 5.4.1.5.1

Simplify each term.

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Step 5.4.1.5.1.1

Simplify each term.

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Step 5.4.1.5.1.1.1

Apply the product rule to $C B$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} \cdot (C^{2} B^{2} - 2 (C (x B^{2})) + {(x B)}^{2} + {(D y)}^{2} - {(B D)}^{2})))}}{2 D^{2}}$

Step 5.4.1.5.1.1.2

Apply the product rule to $x B$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} \cdot (C^{2} B^{2} - 2 C x B^{2} + x^{2} B^{2} + {(D y)}^{2} - {(B D)}^{2})))}}{2 D^{2}}$

Step 5.4.1.5.1.1.3

Apply the product rule to $D y$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} \cdot (C^{2} B^{2} - 2 C x B^{2} + x^{2} B^{2} + D^{2} y^{2} - {(B D)}^{2})))}}{2 D^{2}}$

Step 5.4.1.5.1.1.4

Apply the product rule to $B D$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} \cdot (C^{2} B^{2} - 2 C x B^{2} + x^{2} B^{2} + D^{2} y^{2} - B^{2} D^{2})))}}{2 D^{2}}$

Step 5.4.1.5.1.2

Apply the distributive property.

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} (C^{2} B^{2}) + D^{2} (- 2 C x B^{2}) + D^{2} (x^{2} B^{2}) + D^{2} (D^{2} y^{2}) + D^{2} (- B^{2} D^{2})))}}{2 D^{2}}$

Step 5.4.1.5.1.3

Simplify.

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Step 5.4.1.5.1.3.1

Rewrite using the commutative property of multiplication.

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} C^{2} B^{2} - 2 D^{2} C x B^{2} + D^{2} (x^{2} B^{2}) + D^{2} (D^{2} y^{2}) + D^{2} (- B^{2} D^{2})))}}{2 D^{2}}$

Step 5.4.1.5.1.3.2

Multiply $D^{2}$ by $D^{2}$ by adding the exponents.

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Step 5.4.1.5.1.3.2.1

Move $D^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} C^{2} B^{2} - 2 D^{2} C x B^{2} + D^{2} x^{2} B^{2} + D^{2} D^{2} y^{2} + D^{2} (- B^{2} D^{2})))}}{2 D^{2}}$

Step 5.4.1.5.1.3.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} C^{2} B^{2} - 2 D^{2} C x B^{2} + D^{2} x^{2} B^{2} + D^{2 + 2} y^{2} + D^{2} (- B^{2} D^{2})))}}{2 D^{2}}$

Step 5.4.1.5.1.3.2.3

Add $2$ and $2$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} C^{2} B^{2} - 2 D^{2} C x B^{2} + D^{2} x^{2} B^{2} + D^{4} y^{2} + D^{2} (- B^{2} D^{2})))}}{2 D^{2}}$

Step 5.4.1.5.1.3.3

Rewrite using the commutative property of multiplication.

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} C^{2} B^{2} - 2 D^{2} C x B^{2} + D^{2} x^{2} B^{2} + D^{4} y^{2} - D^{2} (B^{2} D^{2})))}}{2 D^{2}}$

Step 5.4.1.5.1.4

Multiply $D^{2}$ by $D^{2}$ by adding the exponents.

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Step 5.4.1.5.1.4.1

Move $D^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} C^{2} B^{2} - 2 D^{2} C x B^{2} + D^{2} x^{2} B^{2} + D^{4} y^{2} - D^{2} D^{2} B^{2}))}}{2 D^{2}}$

Step 5.4.1.5.1.4.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} C^{2} B^{2} - 2 D^{2} C x B^{2} + D^{2} x^{2} B^{2} + D^{4} y^{2} - D^{2 + 2} B^{2}))}}{2 D^{2}}$

Step 5.4.1.5.1.4.3

Add $2$ and $2$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} C^{2} B^{2} - 2 D^{2} C x B^{2} + D^{2} x^{2} B^{2} + D^{4} y^{2} - D^{4} B^{2}))}}{2 D^{2}}$

Step 5.4.1.5.1.5

Apply the distributive property.

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - (D^{2} C^{2} B^{2}) - (- 2 D^{2} C x B^{2}) - (D^{2} x^{2} B^{2}) - (D^{4} y^{2}) - (- D^{4} B^{2}))}}{2 D^{2}}$

Step 5.4.1.5.1.6

Simplify.

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Step 5.4.1.5.1.6.1

Multiply $- 2$ by $- 1$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - D^{2} C^{2} B^{2} + 2 (D^{2} C x B^{2}) - (D^{2} x^{2} B^{2}) - (D^{4} y^{2}) - (- D^{4} B^{2}))}}{2 D^{2}}$

Step 5.4.1.5.1.6.2

Multiply $- (- D^{4} B^{2})$ .

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Step 5.4.1.5.1.6.2.1

Multiply $- 1$ by $- 1$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - D^{2} C^{2} B^{2} + 2 (D^{2} C x B^{2}) - D^{2} x^{2} B^{2} - D^{4} y^{2} + 1 (D^{4} B^{2}))}}{2 D^{2}}$

Step 5.4.1.5.1.6.2.2

Multiply $D^{4}$ by $1$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - D^{2} C^{2} B^{2} + 2 (D^{2} C x B^{2}) - D^{2} x^{2} B^{2} - D^{4} y^{2} + D^{4} B^{2})}}{2 D^{2}}$

Step 5.4.1.5.1.7

Remove parentheses.

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{4} y^{2} - D^{2} C^{2} B^{2} + 2 D^{2} C x B^{2} - D^{2} x^{2} B^{2} - D^{4} y^{2} + D^{4} B^{2})}}{2 D^{2}}$

Step 5.4.1.5.2

Combine the opposite terms in $D^{4} y^{2} - D^{2} C^{2} B^{2} + 2 D^{2} C x B^{2} - D^{2} x^{2} B^{2} - D^{4} y^{2} + D^{4} B^{2}$ .

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Step 5.4.1.5.2.1

Subtract $D^{4} y^{2}$ from $D^{4} y^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (- D^{2} C^{2} B^{2} + 2 D^{2} C x B^{2} - D^{2} x^{2} B^{2} + 0 + D^{4} B^{2})}}{2 D^{2}}$

Step 5.4.1.5.2.2

Add $- D^{2} C^{2} B^{2} + 2 D^{2} C x B^{2} - D^{2} x^{2} B^{2}$ and $0$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (- D^{2} C^{2} B^{2} + 2 D^{2} C x B^{2} - D^{2} x^{2} B^{2} + D^{4} B^{2})}}{2 D^{2}}$

Step 5.4.1.6

Factor $D^{2} B^{2}$ out of $- D^{2} C^{2} B^{2} + 2 D^{2} C x B^{2} - D^{2} x^{2} B^{2} + D^{4} B^{2}$ .

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Step 5.4.1.6.1

Factor $D^{2} B^{2}$ out of $- D^{2} C^{2} B^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{2} B^{2} (- 1 C^{2}) + 2 D^{2} C x B^{2} - D^{2} x^{2} B^{2} + D^{4} B^{2})}}{2 D^{2}}$

Step 5.4.1.6.2

Factor $D^{2} B^{2}$ out of $2 D^{2} C x B^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{2} B^{2} (- 1 C^{2}) + D^{2} B^{2} (2 C x) - D^{2} x^{2} B^{2} + D^{4} B^{2})}}{2 D^{2}}$

Step 5.4.1.6.3

Factor $D^{2} B^{2}$ out of $- D^{2} x^{2} B^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{2} B^{2} (- 1 C^{2}) + D^{2} B^{2} (2 C x) + D^{2} B^{2} (- 1 x^{2}) + D^{4} B^{2})}}{2 D^{2}}$

Step 5.4.1.6.4

Factor $D^{2} B^{2}$ out of $D^{4} B^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{2} B^{2} (- 1 C^{2}) + D^{2} B^{2} (2 C x) + D^{2} B^{2} (- 1 x^{2}) + D^{2} B^{2} (D^{2}))}}{2 D^{2}}$

Step 5.4.1.6.5

Factor $D^{2} B^{2}$ out of $D^{2} B^{2} (- 1 C^{2}) + D^{2} B^{2} (2 C x)$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{2} B^{2} (- 1 C^{2} + 2 C x) + D^{2} B^{2} (- 1 x^{2}) + D^{2} B^{2} (D^{2}))}}{2 D^{2}}$

Step 5.4.1.6.6

Factor $D^{2} B^{2}$ out of $D^{2} B^{2} (- 1 C^{2} + 2 C x) + D^{2} B^{2} (- 1 x^{2})$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{2} B^{2} (- 1 C^{2} + 2 C x - 1 x^{2}) + D^{2} B^{2} (D^{2}))}}{2 D^{2}}$

Step 5.4.1.6.7

Factor $D^{2} B^{2}$ out of $D^{2} B^{2} (- 1 C^{2} + 2 C x - 1 x^{2}) + D^{2} B^{2} (D^{2})$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{2} B^{2} (- 1 C^{2} + 2 C x - 1 x^{2} + D^{2}))}}{2 D^{2}}$

Step 5.4.1.7

Rewrite $C^{2} - 2 C x + x^{2}$ as ${(C - x)}^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{2} B^{2} (- {(C - x)}^{2} + D^{2}))}}{2 D^{2}}$

Step 5.4.1.8

Reorder $- {(C - x)}^{2}$ and $D^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{2} B^{2} (D^{2} - {(C - x)}^{2}))}}{2 D^{2}}$

Step 5.4.1.9

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

$A = \frac{2 D^{2} y \pm \sqrt{4 (D^{2} B^{2} ((D + C - x) (D - (C - x))))}}{2 D^{2}}$

Step 5.4.1.10

Factor.

$A = \frac{2 D^{2} y \pm \sqrt{4 D^{2} B^{2} (D + C - x) (D - C + x)}}{2 D^{2}}$

Step 5.4.1.11

Rewrite $4 D^{2} B^{2} (D + C - x) (D - C + x)$ as ${(2 D B)}^{2} ((D + C - x) (D - C + x))$ .

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Step 5.4.1.11.1

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

$A = \frac{2 D^{2} y \pm \sqrt{2^{2} D^{2} B^{2} (D + C - x) (D - C + x)}}{2 D^{2}}$

Step 5.4.1.11.2

Rewrite $2^{2} D^{2} B^{2}$ as ${(2 D B)}^{2}$ .

$A = \frac{2 D^{2} y \pm \sqrt{{(2 D B)}^{2} (D + C - x) (D - C + x)}}{2 D^{2}}$

Step 5.4.1.11.3

Add parentheses.

$A = \frac{2 D^{2} y \pm \sqrt{{(2 D B)}^{2} ((D + C - x) (D - C + x))}}{2 D^{2}}$

Step 5.4.1.12

Pull terms out from under the radical.

$A = \frac{2 D^{2} y \pm 2 D B \sqrt{(D + C - x) (D - C + x)}}{2 D^{2}}$

Step 5.4.2

Simplify $\frac{2 D^{2} y \pm 2 D B \sqrt{(D + C - x) (D - C + x)}}{2 D^{2}}$ .

$A = \frac{D y \pm B \sqrt{(D + C - x) (D - C + x)}}{D}$

Step 5.5

The final answer is the combination of both solutions.

$A = \frac{D y + B \sqrt{(D + C - x) (D - C + x)}}{D}$

$A = \frac{D y - B \sqrt{(D + C - x) (D - C + x)}}{D}$

$A = \frac{D y + B \sqrt{(D + C - x) (D - C + x)}}{D}$

$A = \frac{D y - B \sqrt{(D + C - x) (D - C + x)}}{D}$