Precalculus Examples

$25 x^{2} + 2 y^{2} = 50$

Step 1

Subtract $25 x^{2}$ from both sides of the equation.

$2 y^{2} = 50 - 25 x^{2}$

Step 2

Divide each term in $2 y^{2} = 50 - 25 x^{2}$ by $2$ and simplify.

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Step 2.1

Divide each term in $2 y^{2} = 50 - 25 x^{2}$ by $2$ .

$\frac{2 y^{2}}{2} = \frac{50}{2} + \frac{- 25 x^{2}}{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $2$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{2 y^{2}}{2} = \frac{50}{2} + \frac{- 25 x^{2}}{2}$

Step 2.2.1.2

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{50}{2} + \frac{- 25 x^{2}}{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Divide $50$ by $2$ .

$y^{2} = 25 + \frac{- 25 x^{2}}{2}$

Step 2.3.1.2

Move the negative in front of the fraction.

$y^{2} = 25 - \frac{25 x^{2}}{2}$

Step 3

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

$y = \pm \sqrt{25 - \frac{25 x^{2}}{2}}$

Step 4

Simplify $\pm \sqrt{25 - \frac{25 x^{2}}{2}}$ .

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Step 4.1

To write $25$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \pm \sqrt{25 \cdot \frac{2}{2} - \frac{25 x^{2}}{2}}$

Step 4.2

Combine $25$ and $\frac{2}{2}$ .

$y = \pm \sqrt{\frac{25 \cdot 2}{2} - \frac{25 x^{2}}{2}}$

Step 4.3

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{25 \cdot 2 - 25 x^{2}}{2}}$

Step 4.4

Factor $25$ out of $25 \cdot 2 - 25 x^{2}$ .

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Step 4.4.1

Factor $25$ out of $25 \cdot 2$ .

$y = \pm \sqrt{\frac{25 (2) - 25 x^{2}}{2}}$

Step 4.4.2

Factor $25$ out of $- 25 x^{2}$ .

$y = \pm \sqrt{\frac{25 (2) + 25 (- x^{2})}{2}}$

Step 4.4.3

Factor $25$ out of $25 (2) + 25 (- x^{2})$ .

$y = \pm \sqrt{\frac{25 (2 - x^{2})}{2}}$

Step 4.5

Rewrite $\sqrt{\frac{25 (2 - x^{2})}{2}}$ as $\frac{\sqrt{25 (2 - x^{2})}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{25 (2 - x^{2})}}{\sqrt{2}}$

Step 4.6

Simplify the numerator.

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Step 4.6.1

Rewrite $25$ as $5^{2}$ .

$y = \pm \frac{\sqrt{5^{2} (2 - x^{2})}}{\sqrt{2}}$

Step 4.6.2

Pull terms out from under the radical.

$y = \pm \frac{5 \sqrt{2 - x^{2}}}{\sqrt{2}}$

Step 4.7

Multiply $\frac{5 \sqrt{2 - x^{2}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = \pm \frac{5 \sqrt{2 - x^{2}}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Step 4.8

Combine and simplify the denominator.

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Step 4.8.1

Multiply $\frac{5 \sqrt{2 - x^{2}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{\sqrt{2} \sqrt{2}}$

Step 4.8.2

Raise $\sqrt{2}$ to the power of $1$ .

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Step 4.8.3

Raise $\sqrt{2}$ to the power of $1$ .

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Step 4.8.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Step 4.8.5

Add $1$ and $1$ .

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{{\sqrt{2}}^{2}}$

Step 4.8.6

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 4.8.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Step 4.8.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Step 4.8.6.3

Combine $\frac{1}{2}$ and $2$ .

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{2^{\frac{2}{2}}}$

Step 4.8.6.4

Cancel the common factor of $2$ .

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Step 4.8.6.4.1

Cancel the common factor.

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{2^{\frac{2}{2}}}$

Step 4.8.6.4.2

Rewrite the expression.

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{2^{1}}$

Step 4.8.6.5

Evaluate the exponent.

$y = \pm \frac{5 \sqrt{2 - x^{2}} \sqrt{2}}{2}$

Step 4.9

Combine using the product rule for radicals.

$y = \pm \frac{5 \sqrt{2 (2 - x^{2})}}{2}$

Step 5

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

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Step 5.1

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

$y = \frac{5 \sqrt{2 (2 - x^{2})}}{2}$

Step 5.2

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

$y = - \frac{5 \sqrt{2 (2 - x^{2})}}{2}$

Step 5.3

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

$y = \frac{5 \sqrt{2 (2 - x^{2})}}{2}$

$y = - \frac{5 \sqrt{2 (2 - x^{2})}}{2}$

$y = \frac{5 \sqrt{2 (2 - x^{2})}}{2}$

$y = - \frac{5 \sqrt{2 (2 - x^{2})}}{2}$

Step 6

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

$2 (2 - x^{2}) \geq 0$

Step 7

Solve for $x$ .

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Step 7.1

Divide each term in $2 (2 - x^{2}) \geq 0$ by $2$ and simplify.

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Step 7.1.1

Divide each term in $2 (2 - x^{2}) \geq 0$ by $2$ .

$\frac{2 (2 - x^{2})}{2} \geq \frac{0}{2}$

Step 7.1.2

Simplify the left side.

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Step 7.1.2.1

Cancel the common factor of $2$ .

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Step 7.1.2.1.1

Cancel the common factor.

$\frac{2 (2 - x^{2})}{2} \geq \frac{0}{2}$

Step 7.1.2.1.2

Divide $2 - x^{2}$ by $1$ .

$2 - x^{2} \geq \frac{0}{2}$

Step 7.1.3

Simplify the right side.

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Step 7.1.3.1

Divide $0$ by $2$ .

$2 - x^{2} \geq 0$

Step 7.2

Subtract $2$ from both sides of the inequality.

$- x^{2} \geq - 2$

Step 7.3

Divide each term in $- x^{2} \geq - 2$ by $- 1$ and simplify.

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Step 7.3.1

Divide each term in $- x^{2} \geq - 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 2}{- 1}$

Step 7.3.2

Simplify the left side.

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Step 7.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 2}{- 1}$

Step 7.3.2.2

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 2}{- 1}$

Step 7.3.3

Simplify the right side.

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Step 7.3.3.1

Divide $- 2$ by $- 1$ .

$x^{2} \leq 2$

Step 7.4

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

$\sqrt{x^{2}} \leq \sqrt{2}$

Step 7.5

Simplify the left side.

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Step 7.5.1

Pull terms out from under the radical.

$| x | \leq \sqrt{2}$

Step 7.6

Write $| x | \leq \sqrt{2}$ as a piecewise.

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Step 7.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 7.6.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq \sqrt{2}$

Step 7.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 7.6.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq \sqrt{2}$

Step 7.6.5

Write as a piecewise.

${\begin{cases} x \leq \sqrt{2} & x \geq 0 \\ - x \leq \sqrt{2} & x < 0 \end{cases}$

Step 7.7

Find the intersection of $x \leq \sqrt{2}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{2}$

Step 7.8

Solve $- x \leq \sqrt{2}$ when $x < 0$ .

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Step 7.8.1

Divide each term in $- x \leq \sqrt{2}$ by $- 1$ and simplify.

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Step 7.8.1.1

Divide each term in $- x \leq \sqrt{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{2}}{- 1}$

Step 7.8.1.2

Simplify the left side.

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Step 7.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{2}}{- 1}$

Step 7.8.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{2}}{- 1}$

Step 7.8.1.3

Simplify the right side.

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Step 7.8.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{2}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{2}$

Step 7.8.1.3.2

Rewrite $- 1 \cdot \sqrt{2}$ as $- \sqrt{2}$ .

$x \geq - \sqrt{2}$

Step 7.8.2

Find the intersection of $x \geq - \sqrt{2}$ and $x < 0$ .

$- \sqrt{2} \leq x < 0$

Step 7.9

Find the union of the solutions.

$- \sqrt{2} \leq x \leq \sqrt{2}$

Step 8

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

Interval Notation:

$[- \sqrt{2}, \sqrt{2}]$

Set-Builder Notation:

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

Step 9

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

Interval Notation:

$[- 5, 5]$

Set-Builder Notation:

${y | - 5 \leq y \leq 5}$

Step 10

Determine the domain and range.

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

Range: $[- 5, 5], {y | - 5 \leq y \leq 5}$

Step 11