Precalculus Examples

$x^{2} + y^{2} = 121$

Step 1

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

$y^{2} = 121 - x^{2}$

Step 2

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

$y = \pm \sqrt{121 - x^{2}}$

Step 3

Simplify $\pm \sqrt{121 - x^{2}}$ .

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Step 3.1

Rewrite $121$ as $11^{2}$ .

$y = \pm \sqrt{11^{2} - x^{2}}$

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 $a = 11$ and $b = x$ .

$y = \pm \sqrt{(11 + x) (11 - x)}$

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.

$y = \sqrt{(11 + x) (11 - x)}$

Step 4.2

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

$y = - \sqrt{(11 + x) (11 - x)}$

Step 4.3

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

$y = \sqrt{(11 + x) (11 - x)}$

$y = - \sqrt{(11 + x) (11 - x)}$

$y = \sqrt{(11 + x) (11 - x)}$

$y = - \sqrt{(11 + x) (11 - x)}$

Step 5

Set the radicand in $\sqrt{(11 + x) (11 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(11 + x) (11 - x) \geq 0$

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 $0$ .

$11 + x = 0$

$11 - x = 0$

Step 6.2

Set $11 + x$ equal to $0$ and solve for $x$ .

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Step 6.2.1

Set $11 + x$ equal to $0$ .

$11 + x = 0$

Step 6.2.2

Subtract $11$ from both sides of the equation.

$x = - 11$

Step 6.3

Set $11 - x$ equal to $0$ and solve for $x$ .

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Step 6.3.1

Set $11 - x$ equal to $0$ .

$11 - x = 0$

Step 6.3.2

Solve $11 - x = 0$ for $x$ .

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Step 6.3.2.1

Subtract $11$ from both sides of the equation.

$- x = - 11$

Step 6.3.2.2

Divide each term in $- x = - 11$ by $- 1$ and simplify.

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Step 6.3.2.2.1

Divide each term in $- x = - 11$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 11}{- 1}$

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.

$\frac{x}{1} = \frac{- 11}{- 1}$

Step 6.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 11}{- 1}$

Step 6.3.2.2.3

Simplify the right side.

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Step 6.3.2.2.3.1

Divide $- 11$ by $- 1$ .

$x = 11$

Step 6.4

The final solution is all the values that make $(11 + x) (11 - x) \geq 0$ true.

$x = - 11, 11$

Step 6.5

Use each root to create test intervals.

$x < - 11$

$- 11 < x < 11$

$x > 11$

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 $x < - 11$ to see if it makes the inequality true.

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Step 6.6.1.1

Choose a value on the interval $x < - 11$ and see if this value makes the original inequality true.

$x = - 14$

Step 6.6.1.2

Replace $x$ with $- 14$ in the original inequality.

$(11 - 14) (11 - (- 14)) \geq 0$

Step 6.6.1.3

The left side $- 75$ 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 $- 11 < x < 11$ to see if it makes the inequality true.

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Step 6.6.2.1

Choose a value on the interval $- 11 < x < 11$ and see if this value makes the original inequality true.

$x = 0$

Step 6.6.2.2

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

$(11 + 0) (11 - (0)) \geq 0$

Step 6.6.2.3

The left side $121$ 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 $x > 11$ to see if it makes the inequality true.

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Step 6.6.3.1

Choose a value on the interval $x > 11$ and see if this value makes the original inequality true.

$x = 14$

Step 6.6.3.2

Replace $x$ with $14$ in the original inequality.

$(11 + 14) (11 - (14)) \geq 0$

Step 6.6.3.3

The left side $- 75$ 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.

$x < - 11$ False

$- 11 < x < 11$ True

$x > 11$ False

$x < - 11$ False

$- 11 < x < 11$ True

$x > 11$ False

Step 6.7

The solution consists of all of the true intervals.

$- 11 \leq x \leq 11$

Step 7

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

Interval Notation:

$[- 11, 11]$

Set-Builder Notation:

${x | - 11 \leq x \leq 11}$

Step 8

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[- 11, 11]$

Set-Builder Notation:

${y | - 11 \leq y \leq 11}$

Step 9

Determine the domain and range.

Domain: $[- 11, 11], {x | - 11 \leq x \leq 11}$

Range: $[- 11, 11], {y | - 11 \leq y \leq 11}$

Step 10