Pre-Algebra Examples

$\sqrt{64 - 49 x^{2}}$

Step 1

Find the domain for $y = \sqrt{64 - 49 x^{2}}$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

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Step 1.1

Set the radicand in $\sqrt{(8 + 7 x) (8 - 7 x)}$ greater than or equal to $0$ to find where the expression is defined.

$(8 + 7 x) (8 - 7 x) \geq 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

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

$8 + 7 x = 0$

$8 - 7 x = 0$

Step 1.2.2

Set $8 + 7 x$ equal to $0$ and solve for $x$ .

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Step 1.2.2.1

Set $8 + 7 x$ equal to $0$ .

$8 + 7 x = 0$

Step 1.2.2.2

Solve $8 + 7 x = 0$ for $x$ .

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Step 1.2.2.2.1

Subtract $8$ from both sides of the equation.

$7 x = - 8$

Step 1.2.2.2.2

Divide each term in $7 x = - 8$ by $7$ and simplify.

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Step 1.2.2.2.2.1

Divide each term in $7 x = - 8$ by $7$ .

$\frac{7 x}{7} = \frac{- 8}{7}$

Step 1.2.2.2.2.2

Simplify the left side.

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Step 1.2.2.2.2.2.1

Cancel the common factor of $7$ .

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Step 1.2.2.2.2.2.1.1

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 8}{7}$

Step 1.2.2.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 8}{7}$

Step 1.2.2.2.2.3

Simplify the right side.

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Step 1.2.2.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{8}{7}$

Step 1.2.3

Set $8 - 7 x$ equal to $0$ and solve for $x$ .

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Step 1.2.3.1

Set $8 - 7 x$ equal to $0$ .

$8 - 7 x = 0$

Step 1.2.3.2

Solve $8 - 7 x = 0$ for $x$ .

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Step 1.2.3.2.1

Subtract $8$ from both sides of the equation.

$- 7 x = - 8$

Step 1.2.3.2.2

Divide each term in $- 7 x = - 8$ by $- 7$ and simplify.

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Step 1.2.3.2.2.1

Divide each term in $- 7 x = - 8$ by $- 7$ .

$\frac{- 7 x}{- 7} = \frac{- 8}{- 7}$

Step 1.2.3.2.2.2

Simplify the left side.

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Step 1.2.3.2.2.2.1

Cancel the common factor of $- 7$ .

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Step 1.2.3.2.2.2.1.1

Cancel the common factor.

$\frac{- 7 x}{- 7} = \frac{- 8}{- 7}$

Step 1.2.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 8}{- 7}$

Step 1.2.3.2.2.3

Simplify the right side.

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Step 1.2.3.2.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{8}{7}$

Step 1.2.4

The final solution is all the values that make $(8 + 7 x) (8 - 7 x) \geq 0$ true.

$x = - \frac{8}{7}, \frac{8}{7}$

Step 1.2.5

Use each root to create test intervals.

$x < - \frac{8}{7}$

$- \frac{8}{7} < x < \frac{8}{7}$

$x > \frac{8}{7}$

Step 1.2.6

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 1.2.6.1

Test a value on the interval $x < - \frac{8}{7}$ to see if it makes the inequality true.

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Step 1.2.6.1.1

Choose a value on the interval $x < - \frac{8}{7}$ and see if this value makes the original inequality true.

$x = - 4$

Step 1.2.6.1.2

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

$(8 + 7 (- 4)) (8 - 7 \cdot - 4) \geq 0$

Step 1.2.6.1.3

The left side $- 720$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.6.2

Test a value on the interval $- \frac{8}{7} < x < \frac{8}{7}$ to see if it makes the inequality true.

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Step 1.2.6.2.1

Choose a value on the interval $- \frac{8}{7} < x < \frac{8}{7}$ and see if this value makes the original inequality true.

$x = 0$

Step 1.2.6.2.2

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

$(8 + 7 (0)) (8 - 7 \cdot 0) \geq 0$

Step 1.2.6.2.3

The left side $64$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.6.3

Test a value on the interval $x > \frac{8}{7}$ to see if it makes the inequality true.

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Step 1.2.6.3.1

Choose a value on the interval $x > \frac{8}{7}$ and see if this value makes the original inequality true.

$x = 4$

Step 1.2.6.3.2

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

$(8 + 7 (4)) (8 - 7 \cdot 4) \geq 0$

Step 1.2.6.3.3

The left side $- 720$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.6.4

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

$x < - \frac{8}{7}$ False

$- \frac{8}{7} < x < \frac{8}{7}$ True

$x > \frac{8}{7}$ False

$x < - \frac{8}{7}$ False

$- \frac{8}{7} < x < \frac{8}{7}$ True

$x > \frac{8}{7}$ False

Step 1.2.7

The solution consists of all of the true intervals.

$- \frac{8}{7} \leq x \leq \frac{8}{7}$

Step 1.3

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

Interval Notation:

$[- \frac{8}{7}, \frac{8}{7}]$

Set-Builder Notation:

${x | - \frac{8}{7} \leq x \leq \frac{8}{7}}$

Interval Notation:

$[- \frac{8}{7}, \frac{8}{7}]$

Set-Builder Notation:

${x | - \frac{8}{7} \leq x \leq \frac{8}{7}}$

Step 2

To find the end points, substitute the bounds of the $x$ values from the domain into $f (x) = \sqrt{(8 + 7 x) (8 - 7 x)}$ .

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Step 2.1

Replace the variable $x$ with $- \frac{8}{7}$ in the expression.

$f (- \frac{8}{7}) = \sqrt{(8 + 7 (- \frac{8}{7})) (8 - 7 (- \frac{8}{7}))}$

Step 2.2

Simplify the result.

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Step 2.2.1

Cancel the common factor of $7$ .

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Step 2.2.1.1

Move the leading negative in $- \frac{8}{7}$ into the numerator.

$f (- \frac{8}{7}) = \sqrt{(8 + 7 (\frac{- 8}{7})) (8 - 7 (- \frac{8}{7}))}$

Step 2.2.1.2

Cancel the common factor.

$f (- \frac{8}{7}) = \sqrt{(8 + 7 (\frac{- 8}{7})) (8 - 7 (- \frac{8}{7}))}$

Step 2.2.1.3

Rewrite the expression.

$f (- \frac{8}{7}) = \sqrt{(8 - 8) (8 - 7 (- \frac{8}{7}))}$

Step 2.2.2

Subtract $8$ from $8$ .

$f (- \frac{8}{7}) = \sqrt{0 (8 - 7 (- \frac{8}{7}))}$

Step 2.2.3

Cancel the common factor of $7$ .

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Step 2.2.3.1

Move the leading negative in $- \frac{8}{7}$ into the numerator.

$f (- \frac{8}{7}) = \sqrt{0 (8 - 7 (\frac{- 8}{7}))}$

Step 2.2.3.2

Factor $7$ out of $- 7$ .

$f (- \frac{8}{7}) = \sqrt{0 (8 + 7 (- 1) (\frac{- 8}{7}))}$

Step 2.2.3.3

Cancel the common factor.

$f (- \frac{8}{7}) = \sqrt{0 (8 + 7 \cdot (- 1 (\frac{- 8}{7})))}$

Step 2.2.3.4

Rewrite the expression.

$f (- \frac{8}{7}) = \sqrt{0 (8 - 1 \cdot - 8)}$

Step 2.2.4

Simplify the expression.

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Step 2.2.4.1

Multiply $- 1$ by $- 8$ .

$f (- \frac{8}{7}) = \sqrt{0 (8 + 8)}$

Step 2.2.4.2

Add $8$ and $8$ .

$f (- \frac{8}{7}) = \sqrt{0 \cdot 16}$

Step 2.2.4.3

Multiply $0$ by $16$ .

$f (- \frac{8}{7}) = \sqrt{0}$

Step 2.2.4.4

Rewrite $0$ as $0^{2}$ .

$f (- \frac{8}{7}) = \sqrt{0^{2}}$

Step 2.2.4.5

Pull terms out from under the radical, assuming positive real numbers.

$f (- \frac{8}{7}) = 0$

Step 2.2.5

The final answer is $0$ .

$0$

Step 2.3

Replace the variable $x$ with $\frac{8}{7}$ in the expression.

$f (\frac{8}{7}) = \sqrt{(8 + 7 (\frac{8}{7})) (8 - 7 (\frac{8}{7}))}$

Step 2.4

Simplify the result.

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Step 2.4.1

Cancel the common factor of $7$ .

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Step 2.4.1.1

Cancel the common factor.

$f (\frac{8}{7}) = \sqrt{(8 + 7 (\frac{8}{7})) (8 - 7 (\frac{8}{7}))}$

Step 2.4.1.2

Rewrite the expression.

$f (\frac{8}{7}) = \sqrt{(8 + 8) (8 - 7 (\frac{8}{7}))}$

Step 2.4.2

Add $8$ and $8$ .

$f (\frac{8}{7}) = \sqrt{16 (8 - 7 (\frac{8}{7}))}$

Step 2.4.3

Cancel the common factor of $7$ .

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Step 2.4.3.1

Factor $7$ out of $- 7$ .

$f (\frac{8}{7}) = \sqrt{16 (8 + 7 (- 1) (\frac{8}{7}))}$

Step 2.4.3.2

Cancel the common factor.

$f (\frac{8}{7}) = \sqrt{16 (8 + 7 \cdot (- 1 (\frac{8}{7})))}$

Step 2.4.3.3

Rewrite the expression.

$f (\frac{8}{7}) = \sqrt{16 (8 - 1 \cdot 8)}$

Step 2.4.4

Simplify the expression.

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Step 2.4.4.1

Multiply $- 1$ by $8$ .

$f (\frac{8}{7}) = \sqrt{16 (8 - 8)}$

Step 2.4.4.2

Subtract $8$ from $8$ .

$f (\frac{8}{7}) = \sqrt{16 \cdot 0}$

Step 2.4.4.3

Multiply $16$ by $0$ .

$f (\frac{8}{7}) = \sqrt{0}$

Step 2.4.4.4

Rewrite $0$ as $0^{2}$ .

$f (\frac{8}{7}) = \sqrt{0^{2}}$

Step 2.4.4.5

Pull terms out from under the radical, assuming positive real numbers.

$f (\frac{8}{7}) = 0$

Step 2.4.5

The final answer is $0$ .

$0$

Step 3

The end points are $(- \frac{8}{7}, 0), (\frac{8}{7}, 0)$ .

$(- \frac{8}{7}, 0), (\frac{8}{7}, 0)$

Step 4

The square root can be graphed using the points around the vertex $(- \frac{8}{7}, 0), (\frac{8}{7}, 0),$

$\begin{array}{c} x & y \\ - \frac{8}{7} & 0 \\ \frac{8}{7} & 0 \end{array}$

Step 5