Finite Math Examples

$\sqrt{x + 2} \sqrt{x - 3} > 0$

Step 1

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

${(\sqrt{x + 2} \sqrt{x - 3})}^{2} > 0^{2}$

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

${({(x + 2)}^{\frac{1}{2}} \sqrt{x - 3})}^{2} > 0^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

Apply the product rule to ${(x + 2)}^{\frac{1}{2}} \sqrt{x - 3}$ .

${({(x + 2)}^{\frac{1}{2}})}^{2} {\sqrt{x - 3}}^{2} > 0^{2}$

Step 2.2.1.2

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

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Step 2.2.1.2.1

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

${(x + 2)}^{\frac{1}{2} \cdot 2} {\sqrt{x - 3}}^{2} > 0^{2}$

Step 2.2.1.2.2

Cancel the common factor of $2$ .

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Step 2.2.1.2.2.1

Cancel the common factor.

${(x + 2)}^{\frac{1}{2} \cdot 2} {\sqrt{x - 3}}^{2} > 0^{2}$

Step 2.2.1.2.2.2

Rewrite the expression.

${(x + 2)}^{1} {\sqrt{x - 3}}^{2} > 0^{2}$

Step 2.2.1.3

Simplify.

$(x + 2) {\sqrt{x - 3}}^{2} > 0^{2}$

Step 2.2.1.4

Rewrite ${\sqrt{x - 3}}^{2}$ as $x - 3$ .

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Step 2.2.1.4.1

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

$(x + 2) {({(x - 3)}^{\frac{1}{2}})}^{2} > 0^{2}$

Step 2.2.1.4.2

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

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

Step 2.2.1.4.3

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

$(x + 2) {(x - 3)}^{\frac{2}{2}} > 0^{2}$

Step 2.2.1.4.4

Cancel the common factor of $2$ .

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Step 2.2.1.4.4.1

Cancel the common factor.

$(x + 2) {(x - 3)}^{\frac{2}{2}} > 0^{2}$

Step 2.2.1.4.4.2

Rewrite the expression.

$(x + 2) {(x - 3)}^{1} > 0^{2}$

Step 2.2.1.4.5

Simplify.

$(x + 2) (x - 3) > 0^{2}$

Step 2.2.1.5

Expand $(x + 2) (x - 3)$ using the FOIL Method.

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Step 2.2.1.5.1

Apply the distributive property.

$x (x - 3) + 2 (x - 3) > 0^{2}$

Step 2.2.1.5.2

Apply the distributive property.

$x \cdot x + x \cdot - 3 + 2 (x - 3) > 0^{2}$

Step 2.2.1.5.3

Apply the distributive property.

$x \cdot x + x \cdot - 3 + 2 x + 2 \cdot - 3 > 0^{2}$

Step 2.2.1.6

Simplify and combine like terms.

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Step 2.2.1.6.1

Simplify each term.

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Step 2.2.1.6.1.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 3 + 2 x + 2 \cdot - 3 > 0^{2}$

Step 2.2.1.6.1.2

Move $- 3$ to the left of $x$ .

$x^{2} - 3 \cdot x + 2 x + 2 \cdot - 3 > 0^{2}$

Step 2.2.1.6.1.3

Multiply $2$ by $- 3$ .

$x^{2} - 3 x + 2 x - 6 > 0^{2}$

Step 2.2.1.6.2

Add $- 3 x$ and $2 x$ .

$x^{2} - x - 6 > 0^{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

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

$x^{2} - x - 6 > 0$

Step 3

Solve for $x$ .

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Step 3.1

Convert the inequality to an equation.

$x^{2} - x - 6 = 0$

Step 3.2

Factor $x^{2} - x - 6$ using the AC method.

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Step 3.2.1

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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Step 3.2.2

Write the factored form using these integers.

$(x - 3) (x + 2) = 0$

Step 3.3

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

$x - 3 = 0$

$x + 2 = 0$

Step 3.4

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

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Step 3.4.1

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

$x - 3 = 0$

Step 3.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 3.5

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

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Step 3.5.1

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

$x + 2 = 0$

Step 3.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3.6

The final solution is all the values that make $(x - 3) (x + 2) = 0$ true.

$x = 3, - 2$

Step 4

Find the domain of $\sqrt{x + 2} \sqrt{x - 3}$ .

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Step 4.1

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

$x + 2 \geq 0$

Step 4.2

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

Step 4.3

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

$x - 3 \geq 0$

Step 4.4

Add $3$ to both sides of the inequality.

$x \geq 3$

Step 4.5

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

$[3, \infty)$

Step 5

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 3$

$x > 3$

Step 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 6.1

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

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Step 6.1.1

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

$x = - 4$

Step 6.1.2

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

$\sqrt{(- 4) + 2} \sqrt{(- 4) - 3} > 0$

Step 6.1.3

The left side $3.74165738$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 6.2

Test a value on the interval $- 2 < x < 3$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $- 2 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 6.2.2

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

$\sqrt{(0) + 2} \sqrt{(0) - 3} > 0$

Step 6.2.3

The left side is not equal to the right side, which means that the given statement is false.

False

Step 6.3

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

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Step 6.3.1

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

$x = 6$

Step 6.3.2

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

$\sqrt{(6) + 2} \sqrt{(6) - 3} > 0$

Step 6.3.3

The left side $4.89897948$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 6.4

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

$x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

$x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

Step 7

The solution consists of all of the true intervals.

$x < - 2$ or $x > 3$

Step 8

Convert the inequality to interval notation.

$(- \infty, - 2) \cup (3, \infty)$

Step 9