Precalculus Examples

$\frac{63 x^{2} + 81 x}{81 x^{2} - 126 x + 49}$

Step 1

Set the denominator in $\frac{63 x^{2} + 81 x}{81 x^{2} - 126 x + 49}$ equal to $0$ to find where the expression is undefined.

$81 x^{2} - 126 x + 49 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor using the perfect square rule.

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Step 2.1.1

Rewrite $81 x^{2}$ as ${(9 x)}^{2}$ .

${(9 x)}^{2} - 126 x + 49 = 0$

Step 2.1.2

Rewrite $49$ as $7^{2}$ .

${(9 x)}^{2} - 126 x + 7^{2} = 0$

Step 2.1.3

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

$126 x = 2 \cdot (9 x) \cdot 7$

Step 2.1.4

Rewrite the polynomial.

${(9 x)}^{2} - 2 \cdot (9 x) \cdot 7 + 7^{2} = 0$

Step 2.1.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 9 x$ and $b = 7$ .

${(9 x - 7)}^{2} = 0$

Step 2.2

Set the $9 x - 7$ equal to $0$ .

$9 x - 7 = 0$

Step 2.3

Solve for $x$ .

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Step 2.3.1

Add $7$ to both sides of the equation.

$9 x = 7$

Step 2.3.2

Divide each term in $9 x = 7$ by $9$ and simplify.

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Step 2.3.2.1

Divide each term in $9 x = 7$ by $9$ .

$\frac{9 x}{9} = \frac{7}{9}$

Step 2.3.2.2

Simplify the left side.

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Step 2.3.2.2.1

Cancel the common factor of $9$ .

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Step 2.3.2.2.1.1

Cancel the common factor.

$\frac{9 x}{9} = \frac{7}{9}$

Step 2.3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{7}{9}$

Step 3

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

Interval Notation:

$(- \infty, \frac{7}{9}) \cup (\frac{7}{9}, \infty)$

Set-Builder Notation:

${x | x \neq \frac{7}{9}}$

Step 4