Precalculus Examples

$f (x) = \frac{1}{\sqrt{5 x^{2} + 14 x - 3}}$

Step 1

Set the radicand in $\sqrt{5 x^{2} + 14 x - 3}$ greater than or equal to $0$ to find where the expression is defined.

$5 x^{2} + 14 x - 3 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$5 x^{2} + 14 x - 3 = 0$

Step 2.2

Factor by grouping.

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Step 2.2.1

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 $a \cdot c = 5 \cdot - 3 = - 15$ and whose sum is $b = 14$ .

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Step 2.2.1.1

Factor $14$ out of $14 x$ .

$5 x^{2} + 14 (x) - 3 = 0$

Step 2.2.1.2

Rewrite $14$ as $- 1$ plus $15$

$5 x^{2} + (- 1 + 15) x - 3 = 0$

Step 2.2.1.3

Apply the distributive property.

$5 x^{2} - 1 x + 15 x - 3 = 0$

Step 2.2.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1

Group the first two terms and the last two terms.

$(5 x^{2} - 1 x) + 15 x - 3 = 0$

Step 2.2.2.2

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

$x (5 x - 1) + 3 (5 x - 1) = 0$

Step 2.2.3

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

$(5 x - 1) (x + 3) = 0$

Step 2.3

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

$5 x - 1 = 0$

$x + 3 = 0$

Step 2.4

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

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Step 2.4.1

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

$5 x - 1 = 0$

Step 2.4.2

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

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Step 2.4.2.1

Add $1$ to both sides of the equation.

$5 x = 1$

Step 2.4.2.2

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

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Step 2.4.2.2.1

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

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

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $5$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

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

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.5

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.5.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.6

The final solution is all the values that make $(5 x - 1) (x + 3) = 0$ true.

$x = \frac{1}{5}, - 3$

Step 2.7

Use each root to create test intervals.

$x < - 3$

$- 3 < x < \frac{1}{5}$

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

Step 2.8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.8.1

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Step 2.8.1.2

Replace $x$ with $- 6$ in the original inequality.

$5 {(- 6)}^{2} + 14 (- 6) - 3 \geq 0$

Step 2.8.1.3

The left side $93$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.2

Test a value on the interval $- 3 < x < \frac{1}{5}$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $- 3 < x < \frac{1}{5}$ and see if this value makes the original inequality true.

$x = 0$

Step 2.8.2.2

Replace $x$ with $0$ in the original inequality.

$5 {(0)}^{2} + 14 (0) - 3 \geq 0$

Step 2.8.2.3

The left side $- 3$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.8.3

Test a value on the interval $x > \frac{1}{5}$ to see if it makes the inequality true.

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Step 2.8.3.1

Choose a value on the interval $x > \frac{1}{5}$ and see if this value makes the original inequality true.

$x = 3$

Step 2.8.3.2

Replace $x$ with $3$ in the original inequality.

$5 {(3)}^{2} + 14 (3) - 3 \geq 0$

Step 2.8.3.3

The left side $84$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 3$ True

$- 3 < x < \frac{1}{5}$ False

$x > \frac{1}{5}$ True

$x < - 3$ True

$- 3 < x < \frac{1}{5}$ False

$x > \frac{1}{5}$ True

Step 2.9

The solution consists of all of the true intervals.

$x \leq - 3$ or $x \geq \frac{1}{5}$

Step 3

Set the denominator in $\frac{1}{\sqrt{5 x^{2} + 14 x - 3}}$ equal to $0$ to find where the expression is undefined.

$\sqrt{5 x^{2} + 14 x - 3} = 0$

Step 4

Solve for $x$ .

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Step 4.1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{5 x^{2} + 14 x - 3}}^{2} = 0^{2}$

Step 4.2

Simplify each side of the equation.

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Step 4.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 x^{2} + 14 x - 3}$ as ${(5 x^{2} + 14 x - 3)}^{\frac{1}{2}}$ .

${({(5 x^{2} + 14 x - 3)}^{\frac{1}{2}})}^{2} = 0^{2}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Simplify ${({(5 x^{2} + 14 x - 3)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.2.1.1

Multiply the exponents in ${({(5 x^{2} + 14 x - 3)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.2.1.1.1

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

${(5 x^{2} + 14 x - 3)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 4.2.2.1.1.2.1

Cancel the common factor.

${(5 x^{2} + 14 x - 3)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.2.2.1.1.2.2

Rewrite the expression.

${(5 x^{2} + 14 x - 3)}^{1} = 0^{2}$

Step 4.2.2.1.2

Simplify.

$5 x^{2} + 14 x - 3 = 0^{2}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Raising $0$ to any positive power yields $0$ .

$5 x^{2} + 14 x - 3 = 0$

Step 4.3

Solve for $x$ .

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Step 4.3.1

Factor by grouping.

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Step 4.3.1.1

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 $a \cdot c = 5 \cdot - 3 = - 15$ and whose sum is $b = 14$ .

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Step 4.3.1.1.1

Factor $14$ out of $14 x$ .

$5 x^{2} + 14 (x) - 3 = 0$

Step 4.3.1.1.2

Rewrite $14$ as $- 1$ plus $15$

$5 x^{2} + (- 1 + 15) x - 3 = 0$

Step 4.3.1.1.3

Apply the distributive property.

$5 x^{2} - 1 x + 15 x - 3 = 0$

Step 4.3.1.2

Factor out the greatest common factor from each group.

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Step 4.3.1.2.1

Group the first two terms and the last two terms.

$(5 x^{2} - 1 x) + 15 x - 3 = 0$

Step 4.3.1.2.2

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

$x (5 x - 1) + 3 (5 x - 1) = 0$

Step 4.3.1.3

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

$(5 x - 1) (x + 3) = 0$

Step 4.3.2

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

$5 x - 1 = 0$

$x + 3 = 0$

Step 4.3.3

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

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Step 4.3.3.1

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

$5 x - 1 = 0$

Step 4.3.3.2

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

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Step 4.3.3.2.1

Add $1$ to both sides of the equation.

$5 x = 1$

Step 4.3.3.2.2

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

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Step 4.3.3.2.2.1

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

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

Step 4.3.3.2.2.2

Simplify the left side.

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Step 4.3.3.2.2.2.1

Cancel the common factor of $5$ .

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Step 4.3.3.2.2.2.1.1

Cancel the common factor.

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

Step 4.3.3.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 4.3.4

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 4.3.4.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 4.3.4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 4.3.5

The final solution is all the values that make $(5 x - 1) (x + 3) = 0$ true.

$x = \frac{1}{5}, - 3$

Step 4.4

Exclude the solutions that do not make $\sqrt{5 x^{2} + 14 x - 3} = 0$ true.

$x = - 3$

Step 5

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

Interval Notation:

$(- \infty, - 3) \cup [\frac{1}{5}, \infty)$

Set-Builder Notation:

${x | x < - 3, x \geq \frac{1}{5}}$

Step 6