Precalculus Examples

$\frac{\sqrt{x}}{5 x^{2} + 4 x - 1}$

Step 1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 2

Set the denominator in $\frac{\sqrt{x}}{5 x^{2} + 4 x - 1}$ equal to $0$ to find where the expression is undefined.

$5 x^{2} + 4 x - 1 = 0$

Step 3

Solve for $x$ .

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Step 3.1

Factor by grouping.

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Step 3.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot - 1 = - 5$ and whose sum is $b = 4$ .

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Step 3.1.1.1

Factor $4$ out of $4 x$ .

$5 x^{2} + 4 (x) - 1 = 0$

Step 3.1.1.2

Rewrite $4$ as $- 1$ plus $5$

$5 x^{2} + (- 1 + 5) x - 1 = 0$

Step 3.1.1.3

Apply the distributive property.

$5 x^{2} - 1 x + 5 x - 1 = 0$

Step 3.1.2

Factor out the greatest common factor from each group.

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Step 3.1.2.1

Group the first two terms and the last two terms.

$(5 x^{2} - 1 x) + 5 x - 1 = 0$

Step 3.1.2.2

Factor out the greatest common factor (GCF) from each group.

$x (5 x - 1) + 1 (5 x - 1) = 0$

Step 3.1.3

Factor the polynomial by factoring out the greatest common factor, $5 x - 1$ .

$(5 x - 1) (x + 1) = 0$

Step 3.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$5 x - 1 = 0$

$x + 1 = 0$

Step 3.3

Set $5 x - 1$ equal to $0$ and solve for $x$ .

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Step 3.3.1

Set $5 x - 1$ equal to $0$ .

$5 x - 1 = 0$

Step 3.3.2

Solve $5 x - 1 = 0$ for $x$ .

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Step 3.3.2.1

Add $1$ to both sides of the equation.

$5 x = 1$

Step 3.3.2.2

Divide each term in $5 x = 1$ by $5$ and simplify.

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Step 3.3.2.2.1

Divide each term in $5 x = 1$ by $5$ .

$\frac{5 x}{5} = \frac{1}{5}$

Step 3.3.2.2.2

Simplify the left side.

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Step 3.3.2.2.2.1

Cancel the common factor of $5$ .

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Step 3.3.2.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{1}{5}$

Step 3.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{5}$

Step 3.4

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 3.4.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 3.4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.5

The final solution is all the values that make $(5 x - 1) (x + 1) = 0$ true.

$x = \frac{1}{5}, - 1$

Step 4

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

Interval Notation:

$[0, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$

Set-Builder Notation:

${x | x \geq 0, x \neq \frac{1}{5}}$

Step 5