Precalculus Examples

$\frac{a^{2} - b^{2}}{a^{2} - a + a b - b}$

Step 1

Set the denominator in $\frac{a^{2} - b^{2}}{a^{2} - a + a b - b}$ equal to $0$ to find where the expression is undefined.

$a^{2} - a + a b - b = 0$

Step 2

Solve for $a$ .

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Step 2.1

Use the quadratic formula to find the solutions.

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

Step 2.2

Substitute the values $a = 1$ , $b = - 1 + b$ , and $c = - b$ into the quadratic formula and solve for $a$ .

$\frac{- (- 1 + b) \pm \sqrt{{(- 1 + b)}^{2} - 4 \cdot (1 \cdot (- b))}}{2 \cdot 1}$

Step 2.3

Simplify.

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Step 2.3.1

Simplify the numerator.

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Step 2.3.1.1

Apply the distributive property.

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

Step 2.3.1.2

Multiply $- 1$ by $- 1$ .

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

Step 2.3.1.3

Rewrite ${(- 1 + b)}^{2}$ as $(- 1 + b) (- 1 + b)$ .

$a = \frac{1 - b \pm \sqrt{(- 1 + b) (- 1 + b) - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.3.1.4

Expand $(- 1 + b) (- 1 + b)$ using the FOIL Method.

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Step 2.3.1.4.1

Apply the distributive property.

$a = \frac{1 - b \pm \sqrt{- 1 (- 1 + b) + b (- 1 + b) - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.3.1.4.2

Apply the distributive property.

$a = \frac{1 - b \pm \sqrt{- 1 \cdot - 1 - 1 b + b (- 1 + b) - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.3.1.4.3

Apply the distributive property.

$a = \frac{1 - b \pm \sqrt{- 1 \cdot - 1 - 1 b + b \cdot - 1 + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.3.1.5

Simplify and combine like terms.

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Step 2.3.1.5.1

Simplify each term.

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Step 2.3.1.5.1.1

Multiply $- 1$ by $- 1$ .

$a = \frac{1 - b \pm \sqrt{1 - 1 b + b \cdot - 1 + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.3.1.5.1.2

Rewrite $- 1 b$ as $- b$ .

$a = \frac{1 - b \pm \sqrt{1 - b + b \cdot - 1 + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.3.1.5.1.3

Move $- 1$ to the left of $b$ .

$a = \frac{1 - b \pm \sqrt{1 - b - 1 \cdot b + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.3.1.5.1.4

Rewrite $- 1 b$ as $- b$ .

$a = \frac{1 - b \pm \sqrt{1 - b - b + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.3.1.5.1.5

Multiply $b$ by $b$ .

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

Step 2.3.1.5.2

Subtract $b$ from $- b$ .

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

Step 2.3.1.6

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 2.3.1.6.1

Multiply $- 4$ by $1$ .

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

Step 2.3.1.6.2

Multiply $- 4$ by $- 1$ .

$a = \frac{1 - b \pm \sqrt{1 - 2 b + b^{2} + 4 b}}{2 \cdot 1}$

Step 2.3.1.7

Add $- 2 b$ and $4 b$ .

$a = \frac{1 - b \pm \sqrt{1 + b^{2} + 2 b}}{2 \cdot 1}$

Step 2.3.1.8

Reorder terms.

$a = \frac{1 - b \pm \sqrt{b^{2} + 2 b + 1}}{2 \cdot 1}$

Step 2.3.1.9

Factor using the perfect square rule.

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Step 2.3.1.9.1

Rewrite $1$ as $1^{2}$ .

$a = \frac{1 - b \pm \sqrt{b^{2} + 2 b + 1^{2}}}{2 \cdot 1}$

Step 2.3.1.9.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 b = 2 \cdot b \cdot 1$

Step 2.3.1.9.3

Rewrite the polynomial.

$a = \frac{1 - b \pm \sqrt{b^{2} + 2 \cdot b \cdot 1 + 1^{2}}}{2 \cdot 1}$

Step 2.3.1.9.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = b$ and $b = 1$ .

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

Step 2.3.1.10

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

$a = \frac{1 - b \pm (b + 1)}{2 \cdot 1}$

Step 2.3.2

Multiply $2$ by $1$ .

$a = \frac{1 - b \pm (b + 1)}{2}$

Step 2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.4.1

Simplify the numerator.

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Step 2.4.1.1

Apply the distributive property.

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

Step 2.4.1.2

Multiply $- 1$ by $- 1$ .

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

Step 2.4.1.3

Rewrite ${(- 1 + b)}^{2}$ as $(- 1 + b) (- 1 + b)$ .

$a = \frac{1 - b \pm \sqrt{(- 1 + b) (- 1 + b) - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.4.1.4

Expand $(- 1 + b) (- 1 + b)$ using the FOIL Method.

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Step 2.4.1.4.1

Apply the distributive property.

$a = \frac{1 - b \pm \sqrt{- 1 (- 1 + b) + b (- 1 + b) - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.4.1.4.2

Apply the distributive property.

$a = \frac{1 - b \pm \sqrt{- 1 \cdot - 1 - 1 b + b (- 1 + b) - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.4.1.4.3

Apply the distributive property.

$a = \frac{1 - b \pm \sqrt{- 1 \cdot - 1 - 1 b + b \cdot - 1 + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.4.1.5

Simplify and combine like terms.

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Step 2.4.1.5.1

Simplify each term.

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Step 2.4.1.5.1.1

Multiply $- 1$ by $- 1$ .

$a = \frac{1 - b \pm \sqrt{1 - 1 b + b \cdot - 1 + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.4.1.5.1.2

Rewrite $- 1 b$ as $- b$ .

$a = \frac{1 - b \pm \sqrt{1 - b + b \cdot - 1 + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.4.1.5.1.3

Move $- 1$ to the left of $b$ .

$a = \frac{1 - b \pm \sqrt{1 - b - 1 \cdot b + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.4.1.5.1.4

Rewrite $- 1 b$ as $- b$ .

$a = \frac{1 - b \pm \sqrt{1 - b - b + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.4.1.5.1.5

Multiply $b$ by $b$ .

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

Step 2.4.1.5.2

Subtract $b$ from $- b$ .

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

Step 2.4.1.6

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 2.4.1.6.1

Multiply $- 4$ by $1$ .

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

Step 2.4.1.6.2

Multiply $- 4$ by $- 1$ .

$a = \frac{1 - b \pm \sqrt{1 - 2 b + b^{2} + 4 b}}{2 \cdot 1}$

Step 2.4.1.7

Add $- 2 b$ and $4 b$ .

$a = \frac{1 - b \pm \sqrt{1 + b^{2} + 2 b}}{2 \cdot 1}$

Step 2.4.1.8

Reorder terms.

$a = \frac{1 - b \pm \sqrt{b^{2} + 2 b + 1}}{2 \cdot 1}$

Step 2.4.1.9

Factor using the perfect square rule.

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Step 2.4.1.9.1

Rewrite $1$ as $1^{2}$ .

$a = \frac{1 - b \pm \sqrt{b^{2} + 2 b + 1^{2}}}{2 \cdot 1}$

Step 2.4.1.9.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 b = 2 \cdot b \cdot 1$

Step 2.4.1.9.3

Rewrite the polynomial.

$a = \frac{1 - b \pm \sqrt{b^{2} + 2 \cdot b \cdot 1 + 1^{2}}}{2 \cdot 1}$

Step 2.4.1.9.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = b$ and $b = 1$ .

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

Step 2.4.1.10

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

$a = \frac{1 - b \pm (b + 1)}{2 \cdot 1}$

Step 2.4.2

Multiply $2$ by $1$ .

$a = \frac{1 - b \pm (b + 1)}{2}$

Step 2.4.3

Change the $\pm$ to $+$ .

$a = \frac{1 - b + b + 1}{2}$

Step 2.4.4

Simplify the numerator.

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Step 2.4.4.1

Add $1$ and $1$ .

$a = \frac{- b + b + 2}{2}$

Step 2.4.4.2

Add $- b$ and $b$ .

$a = \frac{0 + 2}{2}$

Step 2.4.4.3

Add $0$ and $2$ .

$a = \frac{2}{2}$

Step 2.4.5

Divide $2$ by $2$ .

$a = 1$

Step 2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.5.1

Simplify the numerator.

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Step 2.5.1.1

Apply the distributive property.

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

Step 2.5.1.2

Multiply $- 1$ by $- 1$ .

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

Step 2.5.1.3

Rewrite ${(- 1 + b)}^{2}$ as $(- 1 + b) (- 1 + b)$ .

$a = \frac{1 - b \pm \sqrt{(- 1 + b) (- 1 + b) - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.5.1.4

Expand $(- 1 + b) (- 1 + b)$ using the FOIL Method.

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Step 2.5.1.4.1

Apply the distributive property.

$a = \frac{1 - b \pm \sqrt{- 1 (- 1 + b) + b (- 1 + b) - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.5.1.4.2

Apply the distributive property.

$a = \frac{1 - b \pm \sqrt{- 1 \cdot - 1 - 1 b + b (- 1 + b) - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.5.1.4.3

Apply the distributive property.

$a = \frac{1 - b \pm \sqrt{- 1 \cdot - 1 - 1 b + b \cdot - 1 + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.5.1.5

Simplify and combine like terms.

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Step 2.5.1.5.1

Simplify each term.

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Step 2.5.1.5.1.1

Multiply $- 1$ by $- 1$ .

$a = \frac{1 - b \pm \sqrt{1 - 1 b + b \cdot - 1 + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.5.1.5.1.2

Rewrite $- 1 b$ as $- b$ .

$a = \frac{1 - b \pm \sqrt{1 - b + b \cdot - 1 + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.5.1.5.1.3

Move $- 1$ to the left of $b$ .

$a = \frac{1 - b \pm \sqrt{1 - b - 1 \cdot b + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.5.1.5.1.4

Rewrite $- 1 b$ as $- b$ .

$a = \frac{1 - b \pm \sqrt{1 - b - b + b \cdot b - 4 \cdot 1 \cdot (- 1 b)}}{2 \cdot 1}$

Step 2.5.1.5.1.5

Multiply $b$ by $b$ .

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

Step 2.5.1.5.2

Subtract $b$ from $- b$ .

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

Step 2.5.1.6

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 2.5.1.6.1

Multiply $- 4$ by $1$ .

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

Step 2.5.1.6.2

Multiply $- 4$ by $- 1$ .

$a = \frac{1 - b \pm \sqrt{1 - 2 b + b^{2} + 4 b}}{2 \cdot 1}$

Step 2.5.1.7

Add $- 2 b$ and $4 b$ .

$a = \frac{1 - b \pm \sqrt{1 + b^{2} + 2 b}}{2 \cdot 1}$

Step 2.5.1.8

Reorder terms.

$a = \frac{1 - b \pm \sqrt{b^{2} + 2 b + 1}}{2 \cdot 1}$

Step 2.5.1.9

Factor using the perfect square rule.

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Step 2.5.1.9.1

Rewrite $1$ as $1^{2}$ .

$a = \frac{1 - b \pm \sqrt{b^{2} + 2 b + 1^{2}}}{2 \cdot 1}$

Step 2.5.1.9.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 b = 2 \cdot b \cdot 1$

Step 2.5.1.9.3

Rewrite the polynomial.

$a = \frac{1 - b \pm \sqrt{b^{2} + 2 \cdot b \cdot 1 + 1^{2}}}{2 \cdot 1}$

Step 2.5.1.9.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = b$ and $b = 1$ .

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

Step 2.5.1.10

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

$a = \frac{1 - b \pm (b + 1)}{2 \cdot 1}$

Step 2.5.2

Multiply $2$ by $1$ .

$a = \frac{1 - b \pm (b + 1)}{2}$

Step 2.5.3

Change the $\pm$ to $-$ .

$a = \frac{1 - b - (b + 1)}{2}$

Step 2.5.4

Simplify the numerator.

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Step 2.5.4.1

Apply the distributive property.

$a = \frac{1 - b - b - 1 \cdot 1}{2}$

Step 2.5.4.2

Multiply $- 1$ by $1$ .

$a = \frac{1 - b - b - 1}{2}$

Step 2.5.4.3

Subtract $1$ from $1$ .

$a = \frac{- b - b + 0}{2}$

Step 2.5.4.4

Add $- b - b$ and $0$ .

$a = \frac{- b - b}{2}$

Step 2.5.4.5

Subtract $b$ from $- b$ .

$a = \frac{- 2 b}{2}$

Step 2.5.5

Cancel the common factor of $- 2$ and $2$ .

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Step 2.5.5.1

Factor $2$ out of $- 2 b$ .

$a = \frac{2 (- b)}{2}$

Step 2.5.5.2

Cancel the common factors.

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Step 2.5.5.2.1

Factor $2$ out of $2$ .

$a = \frac{2 (- b)}{2 (1)}$

Step 2.5.5.2.2

Cancel the common factor.

$a = \frac{2 (- b)}{2 \cdot 1}$

Step 2.5.5.2.3

Rewrite the expression.

$a = \frac{- b}{1}$

Step 2.5.5.2.4

Divide $- b$ by $1$ .

$a = - b$

Step 2.6

The final answer is the combination of both solutions.

$a = 1$

$a = - b$

$a = 1$

$a = - b$

Step 3

The domain is all values of $a$ that make the expression defined.

Interval Notation:

$(- \infty, 1) \cup (1, \infty)$

Set-Builder Notation:

${a | a \neq 1}$