Algebra Examples

$\sqrt{x^{2} - x - 2} < \sqrt{x^{2} + 3 x + 2}$

Step 1

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

${\sqrt{x^{2} - x - 2}}^{2} < {\sqrt{x^{2} + 3 x + 2}}^{2}$

Step 2

Simplify each side of the inequality.

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Step 2.1

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

${({(x^{2} - x - 2)}^{\frac{1}{2}})}^{2} < {\sqrt{x^{2} + 3 x + 2}}^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

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

${(x^{2} - x - 2)}^{\frac{1}{2} \cdot 2} < {\sqrt{x^{2} + 3 x + 2}}^{2}$

Step 2.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.2.1.1.2.1

Cancel the common factor.

${(x^{2} - x - 2)}^{\frac{1}{2} \cdot 2} < {\sqrt{x^{2} + 3 x + 2}}^{2}$

Step 2.2.1.1.2.2

Rewrite the expression.

${(x^{2} - x - 2)}^{1} < {\sqrt{x^{2} + 3 x + 2}}^{2}$

Step 2.2.1.2

Simplify.

$x^{2} - x - 2 < {\sqrt{x^{2} + 3 x + 2}}^{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify ${\sqrt{x^{2} + 3 x + 2}}^{2}$ .

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Step 2.3.1.1

Factor $x^{2} + 3 x + 2$ using the AC method.

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Step 2.3.1.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $2$ and whose sum is $3$ .

$1, 2$

Step 2.3.1.1.2

Write the factored form using these integers.

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

Step 2.3.1.2

Rewrite ${\sqrt{(x + 1) (x + 2)}}^{2}$ as $(x + 1) (x + 2)$ .

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Step 2.3.1.2.1

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

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

Step 2.3.1.2.2

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

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

Step 2.3.1.2.3

Combine $\frac{1}{2}$ and $2$ .

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

Step 2.3.1.2.4

Cancel the common factor of $2$ .

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Step 2.3.1.2.4.1

Cancel the common factor.

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

Step 2.3.1.2.4.2

Rewrite the expression.

$x^{2} - x - 2 < {((x + 1) (x + 2))}^{1}$

Step 2.3.1.2.5

Simplify.

$x^{2} - x - 2 < (x + 1) (x + 2)$

Step 2.3.1.3

Expand $(x + 1) (x + 2)$ using the FOIL Method.

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Step 2.3.1.3.1

Apply the distributive property.

$x^{2} - x - 2 < x (x + 2) + 1 (x + 2)$

Step 2.3.1.3.2

Apply the distributive property.

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

Step 2.3.1.3.3

Apply the distributive property.

$x^{2} - x - 2 < x \cdot x + x \cdot 2 + 1 x + 1 \cdot 2$

Step 2.3.1.4

Simplify and combine like terms.

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Step 2.3.1.4.1

Simplify each term.

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Step 2.3.1.4.1.1

Multiply $x$ by $x$ .

$x^{2} - x - 2 < x^{2} + x \cdot 2 + 1 x + 1 \cdot 2$

Step 2.3.1.4.1.2

Move $2$ to the left of $x$ .

$x^{2} - x - 2 < x^{2} + 2 \cdot x + 1 x + 1 \cdot 2$

Step 2.3.1.4.1.3

Multiply $x$ by $1$ .

$x^{2} - x - 2 < x^{2} + 2 x + x + 1 \cdot 2$

Step 2.3.1.4.1.4

Multiply $2$ by $1$ .

$x^{2} - x - 2 < x^{2} + 2 x + x + 2$

Step 2.3.1.4.2

Add $2 x$ and $x$ .

$x^{2} - x - 2 < x^{2} + 3 x + 2$

Step 3

Solve for $x$ .

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Step 3.1

Move all terms containing $x$ to the left side of the inequality.

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Step 3.1.1

Subtract $x^{2}$ from both sides of the inequality.

$x^{2} - x - 2 - x^{2} < 3 x + 2$

Step 3.1.2

Subtract $3 x$ from both sides of the inequality.

$x^{2} - x - 2 - x^{2} - 3 x < 2$

Step 3.1.3

Combine the opposite terms in $x^{2} - x - 2 - x^{2} - 3 x$ .

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Step 3.1.3.1

Subtract $x^{2}$ from $x^{2}$ .

$- x - 2 + 0 - 3 x < 2$

Step 3.1.3.2

Add $- x - 2$ and $0$ .

$- x - 2 - 3 x < 2$

Step 3.1.4

Subtract $3 x$ from $- x$ .

$- 4 x - 2 < 2$

Step 3.2

Move all terms not containing $x$ to the right side of the inequality.

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Step 3.2.1

Add $2$ to both sides of the inequality.

$- 4 x < 2 + 2$

Step 3.2.2

Add $2$ and $2$ .

$- 4 x < 4$

Step 3.3

Divide each term in $- 4 x < 4$ by $- 4$ and simplify.

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Step 3.3.1

Divide each term in $- 4 x < 4$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 4 x}{- 4} > \frac{4}{- 4}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $- 4$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{- 4 x}{- 4} > \frac{4}{- 4}$

Step 3.3.2.1.2

Divide $x$ by $1$ .

$x > \frac{4}{- 4}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $4$ by $- 4$ .

$x > - 1$

Step 4

Find the domain of $\sqrt{(x - 2) (x + 1)} - \sqrt{(x + 1) (x + 2)}$ .

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Step 4.1

Set the radicand in $\sqrt{(x - 2) (x + 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x - 2) (x + 1) \geq 0$

Step 4.2

Solve for $x$ .

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Step 4.2.1

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

$x - 2 = 0$

$x + 1 = 0$

Step 4.2.2

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 4.2.2.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 4.2.2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 4.2.3

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

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Step 4.2.3.1

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

$x + 1 = 0$

Step 4.2.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.2.4

The final solution is all the values that make $(x - 2) (x + 1) \geq 0$ true.

$x = 2, - 1$

Step 4.2.5

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 2$

$x > 2$

Step 4.2.6

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 4.2.6.1

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

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Step 4.2.6.1.1

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 4.2.6.1.2

Replace $x$ with $- 4$ in the original inequality.

$((- 4) - 2) ((- 4) + 1) \geq 0$

Step 4.2.6.1.3

The left side $18$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.2.6.2

Test a value on the interval $- 1 < x < 2$ to see if it makes the inequality true.

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Step 4.2.6.2.1

Choose a value on the interval $- 1 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 4.2.6.2.2

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

$((0) - 2) ((0) + 1) \geq 0$

Step 4.2.6.2.3

The left side $- 2$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.2.6.3

Test a value on the interval $x > 2$ to see if it makes the inequality true.

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Step 4.2.6.3.1

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Step 4.2.6.3.2

Replace $x$ with $4$ in the original inequality.

$((4) - 2) ((4) + 1) \geq 0$

Step 4.2.6.3.3

The left side $10$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.2.6.4

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

$x < - 1$ True

$- 1 < x < 2$ False

$x > 2$ True

$x < - 1$ True

$- 1 < x < 2$ False

$x > 2$ True

Step 4.2.7

The solution consists of all of the true intervals.

$x \leq - 1$ or $x \geq 2$

Step 4.3

Set the radicand in $\sqrt{(x + 1) (x + 2)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 1) (x + 2) \geq 0$

Step 4.4

Solve for $x$ .

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Step 4.4.1

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

$x + 1 = 0$

$x + 2 = 0$

Step 4.4.2

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

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Step 4.4.2.1

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

$x + 1 = 0$

Step 4.4.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.4.3

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 4.4.3.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 4.4.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 4.4.4

The final solution is all the values that make $(x + 1) (x + 2) \geq 0$ true.

$x = - 1, - 2$

Step 4.4.5

Use each root to create test intervals.

$x < - 2$

$- 2 < x < - 1$

$x > - 1$

Step 4.4.6

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 4.4.6.1

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

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Step 4.4.6.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 4.4.6.1.2

Replace $x$ with $- 4$ in the original inequality.

$((- 4) + 1) ((- 4) + 2) \geq 0$

Step 4.4.6.1.3

The left side $6$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.4.6.2

Test a value on the interval $- 2 < x < - 1$ to see if it makes the inequality true.

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Step 4.4.6.2.1

Choose a value on the interval $- 2 < x < - 1$ and see if this value makes the original inequality true.

$x = - 1.5$

Step 4.4.6.2.2

Replace $x$ with $- 1.5$ in the original inequality.

$((- 1.5) + 1) ((- 1.5) + 2) \geq 0$

Step 4.4.6.2.3

The left side $- 0.25$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.4.6.3

Test a value on the interval $x > - 1$ to see if it makes the inequality true.

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Step 4.4.6.3.1

Choose a value on the interval $x > - 1$ and see if this value makes the original inequality true.

$x = 0$

Step 4.4.6.3.2

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

$((0) + 1) ((0) + 2) \geq 0$

Step 4.4.6.3.3

The left side $2$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.4.6.4

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

$x < - 2$ True

$- 2 < x < - 1$ False

$x > - 1$ True

$x < - 2$ True

$- 2 < x < - 1$ False

$x > - 1$ True

Step 4.4.7

The solution consists of all of the true intervals.

$x \leq - 2$ or $x \geq - 1$

Step 4.5

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

$(- \infty, - 2] \cup [- 1, - 1] \cup [2, \infty)$

Step 5

The solution consists of all of the true intervals.

$x > 2$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$x > 2$

Interval Notation:

$(2, \infty)$

Step 7