Finite Math Examples

\sqrt{4 - x} + \sqrt{x^{2} - 9} = y

$\sqrt{4 - x} + \sqrt{x^{2} - 9} = y$

Step 1

Rewrite the equation as

y = \sqrt{4 - x} + \sqrt{x^{2} - 9}

$y = \sqrt{4 - x} + \sqrt{x^{2} - 9}$ .

y = \sqrt{4 - x} + \sqrt{x^{2} - 9}

$y = \sqrt{4 - x} + \sqrt{x^{2} - 9}$

Step 2

Simplify each term.

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Step 2.1

Rewrite

9

$9$ as

3^{2}

$3^{2}$ .

y = \sqrt{4 - x} + \sqrt{x^{2} - 3^{2}}

$y = \sqrt{4 - x} + \sqrt{x^{2} - 3^{2}}$

Step 2.2

Since both terms are perfect squares, factor using the difference of squares formula,

a^{2} - b^{2} = (a + b) (a - b)

$a^{2} - b^{2} = (a + b) (a - b)$ where

a = x

$a = x$ and

b = 3

$b = 3$ .

y = \sqrt{4 - x} + \sqrt{(x + 3) (x - 3)}

$y = \sqrt{4 - x} + \sqrt{(x + 3) (x - 3)}$

y = \sqrt{4 - x} + \sqrt{(x + 3) (x - 3)}

$y = \sqrt{4 - x} + \sqrt{(x + 3) (x - 3)}$

Step 3

Set the radicand in

\sqrt{4 - x}

$\sqrt{4 - x}$ greater than or equal to

0

$0$ to find where the expression is defined.

4 - x \geq 0

$4 - x \geq 0$

Step 4

Solve for

x

$x$ .

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Step 4.1

Subtract

4

$4$ from both sides of the inequality.

- x \geq - 4

$- x \geq - 4$

Step 4.2

Divide each term in

- x \geq - 4

$- x \geq - 4$ by

- 1

$- 1$ and simplify.

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Step 4.2.1

Divide each term in

- x \geq - 4

$- x \geq - 4$ by

- 1

$- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

\frac{- x}{- 1} \leq \frac{- 4}{- 1}

$\frac{- x}{- 1} \leq \frac{- 4}{- 1}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Dividing two negative values results in a positive value.

\frac{x}{1} \leq \frac{- 4}{- 1}

$\frac{x}{1} \leq \frac{- 4}{- 1}$

Step 4.2.2.2

Divide

x

$x$ by

1

$1$ .

x \leq \frac{- 4}{- 1}

$x \leq \frac{- 4}{- 1}$

x \leq \frac{- 4}{- 1}

$x \leq \frac{- 4}{- 1}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide

- 4

$- 4$ by

- 1

$- 1$ .

x \leq 4

$x \leq 4$

x \leq 4

$x \leq 4$

x \leq 4

$x \leq 4$

x \leq 4

$x \leq 4$

Step 5

Set the radicand in

\sqrt{(x + 3) (x - 3)}

$\sqrt{(x + 3) (x - 3)}$ greater than or equal to

0

$0$ to find where the expression is defined.

(x + 3) (x - 3) \geq 0

$(x + 3) (x - 3) \geq 0$

Step 6

Solve for

x

$x$ .

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Step 6.1

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x + 3 = 0

$x + 3 = 0$

x - 3 = 0

$x - 3 = 0$

Step 6.2

Set

x + 3

$x + 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 6.2.1

Set

x + 3

$x + 3$ equal to

0

$0$ .

x + 3 = 0

$x + 3 = 0$

Step 6.2.2

Subtract

3

$3$ from both sides of the equation.

x = - 3

$x = - 3$

x = - 3

$x = - 3$

Step 6.3

Set

x - 3

$x - 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 6.3.1

Set

x - 3

$x - 3$ equal to

0

$0$ .

x - 3 = 0

$x - 3 = 0$

Step 6.3.2

Add

3

$3$ to both sides of the equation.

x = 3

$x = 3$

x = 3

$x = 3$

Step 6.4

The final solution is all the values that make

(x + 3) (x - 3) \geq 0

$(x + 3) (x - 3) \geq 0$ true.

x = - 3, 3

$x = - 3, 3$

Step 6.5

Use each root to create test intervals.

x < - 3

$x < - 3$

- 3 < x < 3

$- 3 < x < 3$

x > 3

$x > 3$

Step 6.6

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 6.6.1

Test a value on the interval

x < - 3

$x < - 3$ to see if it makes the inequality true.

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Step 6.6.1.1

Choose a value on the interval

x < - 3

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

x = - 6

$x = - 6$

Step 6.6.1.2

Replace

x

$x$ with

- 6

$- 6$ in the original inequality.

((- 6) + 3) ((- 6) - 3) \geq 0

$((- 6) + 3) ((- 6) - 3) \geq 0$

Step 6.6.1.3

The left side

27

$27$ is greater than the right side

0

$0$ , which means that the given statement is always true.

True

Step 6.6.2

Test a value on the interval

- 3 < x < 3

$- 3 < x < 3$ to see if it makes the inequality true.

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Step 6.6.2.1

Choose a value on the interval

- 3 < x < 3

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

x = 0

$x = 0$

Step 6.6.2.2

Replace

x

$x$ with

0

$0$ in the original inequality.

((0) + 3) ((0) - 3) \geq 0

$((0) + 3) ((0) - 3) \geq 0$

Step 6.6.2.3

The left side

- 9

$- 9$ is less than the right side

0

$0$ , which means that the given statement is false.

False

Step 6.6.3

Test a value on the interval

x > 3

$x > 3$ to see if it makes the inequality true.

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Step 6.6.3.1

Choose a value on the interval

x > 3

$x > 3$ and see if this value makes the original inequality true.

x = 6

$x = 6$

Step 6.6.3.2

Replace

x

$x$ with

6

$6$ in the original inequality.

((6) + 3) ((6) - 3) \geq 0

$((6) + 3) ((6) - 3) \geq 0$

Step 6.6.3.3

The left side

27

$27$ is greater than the right side

0

$0$ , which means that the given statement is always true.

True

Step 6.6.4

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

x < - 3

$x < - 3$ True

- 3 < x < 3

$- 3 < x < 3$ False

x > 3

$x > 3$ True

x < - 3

$x < - 3$ True

- 3 < x < 3

$- 3 < x < 3$ False

x > 3

$x > 3$ True

Step 6.7

The solution consists of all of the true intervals.

x \leq - 3

$x \leq - 3$ or

x \geq 3

$x \geq 3$

x \leq - 3

$x \leq - 3$ or

$x \geq 3$

Step 7

The domain is all values of

$x$ that make the expression defined.

Interval Notation:

$(- \infty, - 3] \cup [3, 4]$

Set-Builder Notation:

${x | x \leq - 3, 3 \leq x \leq 4}$

Step 8

The range is the set of all valid

$y$ values. Use the graph to find the range.

No solution

Step 9

Determine the domain and range.

No solution

Step 10