Precalculus Examples

$\frac{\sqrt{x^{2} - 1}}{x + 1} \geq 0$

Step 1

Cross multiply.

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Step 1.1

Cross multiply by setting the product of the numerator of the right side and the denominator of the left side equal to the product of the numerator of the left side and the denominator of the right side.

$0 \cdot (x + 1) \geq \sqrt{x^{2} - 1}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Multiply $0$ by $x + 1$ .

$0 \geq \sqrt{x^{2} - 1}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Simplify $\sqrt{x^{2} - 1}$ .

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Step 1.3.1.1

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

$0 \geq \sqrt{x^{2} - 1^{2}}$

Step 1.3.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$0 \geq \sqrt{(x + 1) (x - 1)}$

Step 2

Rewrite so $\sqrt{(x + 1) (x - 1)}$ is on the left side of the inequality.

$\sqrt{(x + 1) (x - 1)} \leq 0$

Step 3

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

${\sqrt{(x + 1) (x - 1)}}^{2} \leq 0^{2}$

Step 4

Simplify each side of the inequality.

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Step 4.1

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

${({((x + 1) (x - 1))}^{\frac{1}{2}})}^{2} \leq 0^{2}$

Step 4.2

Simplify the left side.

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Step 4.2.1

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

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Step 4.2.1.1

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

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Step 4.2.1.1.1

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

${((x + 1) (x - 1))}^{\frac{1}{2} \cdot 2} \leq 0^{2}$

Step 4.2.1.1.2

Cancel the common factor of $2$ .

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Step 4.2.1.1.2.1

Cancel the common factor.

${((x + 1) (x - 1))}^{\frac{1}{2} \cdot 2} \leq 0^{2}$

Step 4.2.1.1.2.2

Rewrite the expression.

${((x + 1) (x - 1))}^{1} \leq 0^{2}$

Step 4.2.1.2

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

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Step 4.2.1.2.1

Apply the distributive property.

${(x (x - 1) + 1 (x - 1))}^{1} \leq 0^{2}$

Step 4.2.1.2.2

Apply the distributive property.

${(x \cdot x + x \cdot - 1 + 1 (x - 1))}^{1} \leq 0^{2}$

Step 4.2.1.2.3

Apply the distributive property.

${(x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1)}^{1} \leq 0^{2}$

Step 4.2.1.3

Simplify and combine like terms.

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Step 4.2.1.3.1

Simplify each term.

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Step 4.2.1.3.1.1

Multiply $x$ by $x$ .

${(x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1)}^{1} \leq 0^{2}$

Step 4.2.1.3.1.2

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

${(x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1)}^{1} \leq 0^{2}$

Step 4.2.1.3.1.3

Rewrite $- 1 x$ as $- x$ .

${(x^{2} - x + 1 x + 1 \cdot - 1)}^{1} \leq 0^{2}$

Step 4.2.1.3.1.4

Multiply $x$ by $1$ .

${(x^{2} - x + x + 1 \cdot - 1)}^{1} \leq 0^{2}$

Step 4.2.1.3.1.5

Multiply $- 1$ by $1$ .

${(x^{2} - x + x - 1)}^{1} \leq 0^{2}$

Step 4.2.1.3.2

Add $- x$ and $x$ .

${(x^{2} + 0 - 1)}^{1} \leq 0^{2}$

Step 4.2.1.3.3

Add $x^{2}$ and $0$ .

${(x^{2} - 1)}^{1} \leq 0^{2}$

Step 4.2.1.4

Simplify.

$x^{2} - 1 \leq 0^{2}$

Step 4.3

Simplify the right side.

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Step 4.3.1

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

$x^{2} - 1 \leq 0$

Step 5

Solve for $x$ .

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Step 5.1

Add $1$ to both sides of the inequality.

$x^{2} \leq 1$

Step 5.2

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} \leq \sqrt{1}$

Step 5.3

Simplify the equation.

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Step 5.3.1

Simplify the left side.

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Step 5.3.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{1}$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Any root of $1$ is $1$ .

$| x | \leq 1$

Step 5.4

Write $| x | \leq 1$ as a piecewise.

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Step 5.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 5.4.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 1$

Step 5.4.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 5.4.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 1$

Step 5.4.5

Write as a piecewise.

${\begin{cases} x \leq 1 & x \geq 0 \\ - x \leq 1 & x < 0 \end{cases}$

Step 5.5

Find the intersection of $x \leq 1$ and $x \geq 0$ .

$0 \leq x \leq 1$

Step 5.6

Solve $- x \leq 1$ when $x < 0$ .

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Step 5.6.1

Divide each term in $- x \leq 1$ by $- 1$ and simplify.

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Step 5.6.1.1

Divide each term in $- x \leq 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{1}{- 1}$

Step 5.6.1.2

Simplify the left side.

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Step 5.6.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{1}{- 1}$

Step 5.6.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{1}{- 1}$

Step 5.6.1.3

Simplify the right side.

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Step 5.6.1.3.1

Divide $1$ by $- 1$ .

$x \geq - 1$

Step 5.6.2

Find the intersection of $x \geq - 1$ and $x < 0$ .

$- 1 \leq x < 0$

Step 5.7

Find the union of the solutions.

$- 1 \leq x \leq 1$

Step 6

Convert the inequality to interval notation.

$[- 1, 1]$

Step 7