Precalculus Examples

$x^{2} + y^{2} + 2 x = 0$

Step 1

Move all terms not containing $y$ to the right side of the equation.

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Step 1.1

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

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

Step 1.2

Subtract $2 x$ from both sides of the equation.

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

Step 2

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

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

Step 3

Factor $x$ out of $- x^{2} - 2 x$ .

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Step 3.1

Factor $x$ out of $- x^{2}$ .

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

Step 3.2

Factor $x$ out of $- 2 x$ .

$y = \pm \sqrt{x (- x) + x \cdot - 2}$

Step 3.3

Factor $x$ out of $x (- x) + x \cdot - 2$ .

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

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.

$y = \sqrt{x (- x - 2)}$

Step 4.2

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

$y = - \sqrt{x (- x - 2)}$

Step 4.3

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

$y = \sqrt{x (- x - 2)}$

$y = - \sqrt{x (- x - 2)}$

$y = \sqrt{x (- x - 2)}$

$y = - \sqrt{x (- x - 2)}$

Step 5

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

$x (- x - 2) \geq 0$

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 $0$ .

$x = 0$

$- x - 2 = 0$

Step 6.2

Set $x$ equal to $0$ .

$x = 0$

Step 6.3

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

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Step 6.3.1

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

$- x - 2 = 0$

Step 6.3.2

Solve $- x - 2 = 0$ for $x$ .

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Step 6.3.2.1

Add $2$ to both sides of the equation.

$- x = 2$

Step 6.3.2.2

Divide each term in $- x = 2$ by $- 1$ and simplify.

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Step 6.3.2.2.1

Divide each term in $- x = 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{2}{- 1}$

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.

$\frac{x}{1} = \frac{2}{- 1}$

Step 6.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{2}{- 1}$

Step 6.3.2.2.3

Simplify the right side.

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Step 6.3.2.2.3.1

Divide $2$ by $- 1$ .

$x = - 2$

Step 6.4

The final solution is all the values that make $x (- x - 2) \geq 0$ true.

$x = 0, - 2$

Step 6.5

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 0$

$x > 0$

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 $x < - 2$ to see if it makes the inequality true.

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Step 6.6.1.1

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

$x = - 4$

Step 6.6.1.2

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

$(- 4) (- (- 4) - 2) \geq 0$

Step 6.6.1.3

The left side $- 8$ 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 $- 2 < x < 0$ to see if it makes the inequality true.

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Step 6.6.2.1

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Step 6.6.2.2

Replace $x$ with $- 1$ in the original inequality.

$(- 1) (- (- 1) - 2) \geq 0$

Step 6.6.2.3

The left side $1$ 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 $x > 0$ to see if it makes the inequality true.

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Step 6.6.3.1

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

$x = 2$

Step 6.6.3.2

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

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

Step 6.6.3.3

The left side $- 8$ 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.

$x < - 2$ False

$- 2 < x < 0$ True

$x > 0$ False

$x < - 2$ False

$- 2 < x < 0$ True

$x > 0$ False

Step 6.7

The solution consists of all of the true intervals.

$- 2 \leq x \leq 0$

Step 7

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

Interval Notation:

$[- 2, 0]$

Set-Builder Notation:

${x | - 2 \leq x \leq 0}$

Step 8

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

Interval Notation:

$[- 1, 1]$

Set-Builder Notation:

${y | - 1 \leq y \leq 1}$

Step 9

Determine the domain and range.

Domain: $[- 2, 0], {x | - 2 \leq x \leq 0}$

Range: $[- 1, 1], {y | - 1 \leq y \leq 1}$

Step 10