Finite Math Examples

${(x - 3)}^{2} + {(y - 6)}^{2} = 9$

Step 1

Subtract ${(x - 3)}^{2}$ from both sides of the equation.

${(y - 6)}^{2} = 9 - {(x - 3)}^{2}$

Step 2

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

$y - 6 = \pm \sqrt{9 - {(x - 3)}^{2}}$

Step 3

Simplify $\pm \sqrt{9 - {(x - 3)}^{2}}$ .

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Step 3.1

Rewrite $9$ as $3^{2}$ .

$y - 6 = \pm \sqrt{3^{2} - {(x - 3)}^{2}}$

Step 3.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = x - 3$ .

$y - 6 = \pm \sqrt{(3 + x - 3) (3 - (x - 3))}$

Step 3.3

Simplify.

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Step 3.3.1

Subtract $3$ from $3$ .

$y - 6 = \pm \sqrt{(x + 0) (3 - (x - 3))}$

Step 3.3.2

Add $x$ and $0$ .

$y - 6 = \pm \sqrt{x (3 - (x - 3))}$

Step 3.3.3

Apply the distributive property.

$y - 6 = \pm \sqrt{x (3 - x - - 3)}$

Step 3.3.4

Multiply $- 1$ by $- 3$ .

$y - 6 = \pm \sqrt{x (3 - x + 3)}$

Step 3.3.5

Add $3$ and $3$ .

$y - 6 = \pm \sqrt{x (- x + 6)}$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.1

First, use the positive value of the $\pm$ to find the first solution.

$y - 6 = \sqrt{x (- x + 6)}$

Step 4.2

Add $6$ to both sides of the equation.

$y = \sqrt{x (- x + 6)} + 6$

Step 4.3

Next, use the negative value of the $\pm$ to find the second solution.

$y - 6 = - \sqrt{x (- x + 6)}$

Step 4.4

Add $6$ to both sides of the equation.

$y = - \sqrt{x (- x + 6)} + 6$

Step 4.5

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{x (- x + 6)} + 6$

$y = - \sqrt{x (- x + 6)} + 6$

$y = \sqrt{x (- x + 6)} + 6$

$y = - \sqrt{x (- x + 6)} + 6$

Step 5

Set the radicand in $\sqrt{x (- x + 6)}$ greater than or equal to $0$ to find where the expression is defined.

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

Step 6

Solve for $x$ .

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Step 6.1

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

$x = 0$

$- x + 6 = 0$

Step 6.2

Set $x$ equal to $0$ .

$x = 0$

Step 6.3

Set $- x + 6$ equal to $0$ and solve for $x$ .

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Step 6.3.1

Set $- x + 6$ equal to $0$ .

$- x + 6 = 0$

Step 6.3.2

Solve $- x + 6 = 0$ for $x$ .

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Step 6.3.2.1

Subtract $6$ from both sides of the equation.

$- x = - 6$

Step 6.3.2.2

Divide each term in $- x = - 6$ by $- 1$ and simplify.

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Step 6.3.2.2.1

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Step 6.3.2.2.2

Simplify the left side.

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Step 6.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Step 6.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Step 6.3.2.2.3

Simplify the right side.

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Step 6.3.2.2.3.1

Divide $- 6$ by $- 1$ .

$x = 6$

Step 6.4

The final solution is all the values that make $x (- x + 6) \geq 0$ true.

$x = 0, 6$

Step 6.5

Use each root to create test intervals.

$x < 0$

$0 < x < 6$

$x > 6$

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 $x < 0$ to see if it makes the inequality true.

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Step 6.6.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 6.6.1.2

Replace $x$ with $- 2$ in the original inequality.

$(- 2) (- (- 2) + 6) \geq 0$

Step 6.6.1.3

The left side $- 16$ is less than the right side $0$ , which means that the given statement is false.

False

Step 6.6.2

Test a value on the interval $0 < x < 6$ to see if it makes the inequality true.

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Step 6.6.2.1

Choose a value on the interval $0 < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Step 6.6.2.2

Replace $x$ with $3$ in the original inequality.

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

Step 6.6.2.3

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 6.6.3

Test a value on the interval $x > 6$ to see if it makes the inequality true.

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Step 6.6.3.1

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 6.6.3.2

Replace $x$ with $8$ in the original inequality.

$(8) (- (8) + 6) \geq 0$

Step 6.6.3.3

The left side $- 16$ is less than the right side $0$ , which means that the given statement is false.

False

Step 6.6.4

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

$x < 0$ False

$0 < x < 6$ True

$x > 6$ False

$x < 0$ False

$0 < x < 6$ True

$x > 6$ False

Step 6.7

The solution consists of all of the true intervals.

$0 \leq x \leq 6$

Step 7

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

Interval Notation:

$[0, 6]$

Set-Builder Notation:

${x | 0 \leq x \leq 6}$

Step 8