Algebra Examples

$\sqrt{10 - x^{2}} + 6 < | 10 - x^{2} |$

Step 1

Write $\sqrt{10 - x^{2}} + 6 < | 10 - x^{2} |$ as a piecewise.

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Step 1.1

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

$10 - x^{2} \geq 0$

Step 1.2

Solve the inequality.

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Step 1.2.1

Subtract $10$ from both sides of the inequality.

$- x^{2} \geq - 10$

Step 1.2.2

Divide each term in $- x^{2} \geq - 10$ by $- 1$ and simplify.

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Step 1.2.2.1

Divide each term in $- x^{2} \geq - 10$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 10}{- 1}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 10}{- 1}$

Step 1.2.2.2.2

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

$x^{2} \leq \frac{- 10}{- 1}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Divide $- 10$ by $- 1$ .

$x^{2} \leq 10$

Step 1.2.3

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

$\sqrt{x^{2}} \leq \sqrt{10}$

Step 1.2.4

Simplify the left side.

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Step 1.2.4.1

Pull terms out from under the radical.

$| x | \leq \sqrt{10}$

Step 1.2.5

Write $| x | \leq \sqrt{10}$ as a piecewise.

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Step 1.2.5.1

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

$x \geq 0$

Step 1.2.5.2

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

$x \leq \sqrt{10}$

Step 1.2.5.3

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

$x < 0$

Step 1.2.5.4

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

$- x \leq \sqrt{10}$

Step 1.2.5.5

Write as a piecewise.

${\begin{cases} x \leq \sqrt{10} & x \geq 0 \\ - x \leq \sqrt{10} & x < 0 \end{cases}$

Step 1.2.6

Find the intersection of $x \leq \sqrt{10}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{10}$

Step 1.2.7

Solve $- x \leq \sqrt{10}$ when $x < 0$ .

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Step 1.2.7.1

Divide each term in $- x \leq \sqrt{10}$ by $- 1$ and simplify.

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Step 1.2.7.1.1

Divide each term in $- x \leq \sqrt{10}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{10}}{- 1}$

Step 1.2.7.1.2

Simplify the left side.

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Step 1.2.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{10}}{- 1}$

Step 1.2.7.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{10}}{- 1}$

Step 1.2.7.1.3

Simplify the right side.

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Step 1.2.7.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{10}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{10}$

Step 1.2.7.1.3.2

Rewrite $- 1 \cdot \sqrt{10}$ as $- \sqrt{10}$ .

$x \geq - \sqrt{10}$

Step 1.2.7.2

Find the intersection of $x \geq - \sqrt{10}$ and $x < 0$ .

$- \sqrt{10} \leq x < 0$

Step 1.2.8

Find the union of the solutions.

$- \sqrt{10} \leq x \leq \sqrt{10}$

Step 1.3

In the piece where $10 - x^{2}$ is non-negative, remove the absolute value.

$\sqrt{10 - x^{2}} + 6 < 10 - x^{2}$

Step 1.4

Find the domain of $\sqrt{10 - x^{2}} + 6 < 10 - x^{2}$ and find the intersection with $- \sqrt{10} \leq x \leq \sqrt{10}$ .

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Step 1.4.1

Find the domain of $\sqrt{10 - x^{2}} + 6 < 10 - x^{2}$ .

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Step 1.4.1.1

Set the radicand in $\sqrt{10 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$10 - x^{2} \geq 0$

Step 1.4.1.2

Solve for $x$ .

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Step 1.4.1.2.1

Subtract $10$ from both sides of the inequality.

$- x^{2} \geq - 10$

Step 1.4.1.2.2

Divide each term in $- x^{2} \geq - 10$ by $- 1$ and simplify.

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Step 1.4.1.2.2.1

Divide each term in $- x^{2} \geq - 10$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 10}{- 1}$

Step 1.4.1.2.2.2

Simplify the left side.

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Step 1.4.1.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 10}{- 1}$

Step 1.4.1.2.2.2.2

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

$x^{2} \leq \frac{- 10}{- 1}$

Step 1.4.1.2.2.3

Simplify the right side.

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Step 1.4.1.2.2.3.1

Divide $- 10$ by $- 1$ .

$x^{2} \leq 10$

Step 1.4.1.2.3

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

$\sqrt{x^{2}} \leq \sqrt{10}$

Step 1.4.1.2.4

Simplify the left side.

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Step 1.4.1.2.4.1

Pull terms out from under the radical.

$| x | \leq \sqrt{10}$

Step 1.4.1.2.5

Write $| x | \leq \sqrt{10}$ as a piecewise.

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Step 1.4.1.2.5.1

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

$x \geq 0$

Step 1.4.1.2.5.2

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

$x \leq \sqrt{10}$

Step 1.4.1.2.5.3

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

$x < 0$

Step 1.4.1.2.5.4

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

$- x \leq \sqrt{10}$

Step 1.4.1.2.5.5

Write as a piecewise.

${\begin{cases} x \leq \sqrt{10} & x \geq 0 \\ - x \leq \sqrt{10} & x < 0 \end{cases}$

Step 1.4.1.2.6

Find the intersection of $x \leq \sqrt{10}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{10}$

Step 1.4.1.2.7

Solve $- x \leq \sqrt{10}$ when $x < 0$ .

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Step 1.4.1.2.7.1

Divide each term in $- x \leq \sqrt{10}$ by $- 1$ and simplify.

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Step 1.4.1.2.7.1.1

Divide each term in $- x \leq \sqrt{10}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{10}}{- 1}$

Step 1.4.1.2.7.1.2

Simplify the left side.

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Step 1.4.1.2.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{10}}{- 1}$

Step 1.4.1.2.7.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{10}}{- 1}$

Step 1.4.1.2.7.1.3

Simplify the right side.

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Step 1.4.1.2.7.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{10}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{10}$

Step 1.4.1.2.7.1.3.2

Rewrite $- 1 \cdot \sqrt{10}$ as $- \sqrt{10}$ .

$x \geq - \sqrt{10}$

Step 1.4.1.2.7.2

Find the intersection of $x \geq - \sqrt{10}$ and $x < 0$ .

$- \sqrt{10} \leq x < 0$

Step 1.4.1.2.8

Find the union of the solutions.

$- \sqrt{10} \leq x \leq \sqrt{10}$

Step 1.4.1.3

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

$[- \sqrt{10}, \sqrt{10}]$

Step 1.4.2

Find the intersection of $- \sqrt{10} \leq x \leq \sqrt{10}$ and $[- \sqrt{10}, \sqrt{10}]$ .

$- \sqrt{10} \leq x \leq \sqrt{10}$

Step 1.5

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

$10 - x^{2} < 0$

Step 1.6

Solve the inequality.

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Step 1.6.1

Subtract $10$ from both sides of the inequality.

$- x^{2} < - 10$

Step 1.6.2

Divide each term in $- x^{2} < - 10$ by $- 1$ and simplify.

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Step 1.6.2.1

Divide each term in $- x^{2} < - 10$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} > \frac{- 10}{- 1}$

Step 1.6.2.2

Simplify the left side.

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Step 1.6.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} > \frac{- 10}{- 1}$

Step 1.6.2.2.2

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

$x^{2} > \frac{- 10}{- 1}$

Step 1.6.2.3

Simplify the right side.

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Step 1.6.2.3.1

Divide $- 10$ by $- 1$ .

$x^{2} > 10$

Step 1.6.3

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

$\sqrt{x^{2}} > \sqrt{10}$

Step 1.6.4

Simplify the left side.

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Step 1.6.4.1

Pull terms out from under the radical.

$| x | > \sqrt{10}$

Step 1.6.5

Write $| x | > \sqrt{10}$ as a piecewise.

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Step 1.6.5.1

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

$x \geq 0$

Step 1.6.5.2

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

$x > \sqrt{10}$

Step 1.6.5.3

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

$x < 0$

Step 1.6.5.4

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

$- x > \sqrt{10}$

Step 1.6.5.5

Write as a piecewise.

${\begin{cases} x > \sqrt{10} & x \geq 0 \\ - x > \sqrt{10} & x < 0 \end{cases}$

Step 1.6.6

Find the intersection of $x > \sqrt{10}$ and $x \geq 0$ .

$x > \sqrt{10}$

Step 1.6.7

Divide each term in $- x > \sqrt{10}$ by $- 1$ and simplify.

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Step 1.6.7.1

Divide each term in $- x > \sqrt{10}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} < \frac{\sqrt{10}}{- 1}$

Step 1.6.7.2

Simplify the left side.

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Step 1.6.7.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} < \frac{\sqrt{10}}{- 1}$

Step 1.6.7.2.2

Divide $x$ by $1$ .

$x < \frac{\sqrt{10}}{- 1}$

Step 1.6.7.3

Simplify the right side.

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Step 1.6.7.3.1

Move the negative one from the denominator of $\frac{\sqrt{10}}{- 1}$ .

$x < - 1 \cdot \sqrt{10}$

Step 1.6.7.3.2

Rewrite $- 1 \cdot \sqrt{10}$ as $- \sqrt{10}$ .

$x < - \sqrt{10}$

Step 1.6.8

Find the union of the solutions.

$x < - \sqrt{10}$ or $x > \sqrt{10}$

Step 1.7

In the piece where $10 - x^{2}$ is negative, remove the absolute value and multiply by $- 1$ .

$\sqrt{10 - x^{2}} + 6 < - (10 - x^{2})$

Step 1.8

Find the domain of $\sqrt{10 - x^{2}} + 6 < - (10 - x^{2})$ and find the intersection with $x < - \sqrt{10} or x > \sqrt{10}$ .

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Step 1.8.1

Find the domain of $\sqrt{10 - x^{2}} + 6 < - (10 - x^{2})$ .

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Step 1.8.1.1

Set the radicand in $\sqrt{10 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$10 - x^{2} \geq 0$

Step 1.8.1.2

Solve for $x$ .

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Step 1.8.1.2.1

Subtract $10$ from both sides of the inequality.

$- x^{2} \geq - 10$

Step 1.8.1.2.2

Divide each term in $- x^{2} \geq - 10$ by $- 1$ and simplify.

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Step 1.8.1.2.2.1

Divide each term in $- x^{2} \geq - 10$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 10}{- 1}$

Step 1.8.1.2.2.2

Simplify the left side.

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Step 1.8.1.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 10}{- 1}$

Step 1.8.1.2.2.2.2

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

$x^{2} \leq \frac{- 10}{- 1}$

Step 1.8.1.2.2.3

Simplify the right side.

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Step 1.8.1.2.2.3.1

Divide $- 10$ by $- 1$ .

$x^{2} \leq 10$

Step 1.8.1.2.3

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

$\sqrt{x^{2}} \leq \sqrt{10}$

Step 1.8.1.2.4

Simplify the left side.

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Step 1.8.1.2.4.1

Pull terms out from under the radical.

$| x | \leq \sqrt{10}$

Step 1.8.1.2.5

Write $| x | \leq \sqrt{10}$ as a piecewise.

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Step 1.8.1.2.5.1

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

$x \geq 0$

Step 1.8.1.2.5.2

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

$x \leq \sqrt{10}$

Step 1.8.1.2.5.3

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

$x < 0$

Step 1.8.1.2.5.4

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

$- x \leq \sqrt{10}$

Step 1.8.1.2.5.5

Write as a piecewise.

${\begin{cases} x \leq \sqrt{10} & x \geq 0 \\ - x \leq \sqrt{10} & x < 0 \end{cases}$

Step 1.8.1.2.6

Find the intersection of $x \leq \sqrt{10}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{10}$

Step 1.8.1.2.7

Solve $- x \leq \sqrt{10}$ when $x < 0$ .

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Step 1.8.1.2.7.1

Divide each term in $- x \leq \sqrt{10}$ by $- 1$ and simplify.

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Step 1.8.1.2.7.1.1

Divide each term in $- x \leq \sqrt{10}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{10}}{- 1}$

Step 1.8.1.2.7.1.2

Simplify the left side.

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Step 1.8.1.2.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{10}}{- 1}$

Step 1.8.1.2.7.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{10}}{- 1}$

Step 1.8.1.2.7.1.3

Simplify the right side.

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Step 1.8.1.2.7.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{10}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{10}$

Step 1.8.1.2.7.1.3.2

Rewrite $- 1 \cdot \sqrt{10}$ as $- \sqrt{10}$ .

$x \geq - \sqrt{10}$

Step 1.8.1.2.7.2

Find the intersection of $x \geq - \sqrt{10}$ and $x < 0$ .

$- \sqrt{10} \leq x < 0$

Step 1.8.1.2.8

Find the union of the solutions.

$- \sqrt{10} \leq x \leq \sqrt{10}$

Step 1.8.1.3

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

$[- \sqrt{10}, \sqrt{10}]$

Step 1.8.2

Find the intersection of $x < - \sqrt{10} or x > \sqrt{10}$ and $[- \sqrt{10}, \sqrt{10}]$ .

$No solution$

Step 1.9

Write as a piecewise.

${\begin{cases} \sqrt{10 - x^{2}} + 6 < 10 - x^{2} & - \sqrt{10} \leq x \leq \sqrt{10} \end{cases}$

Step 2

Solve $\sqrt{10 - x^{2}} + 6 < 10 - x^{2}$ for $x$ .

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Step 2.1

Move all terms not containing $\sqrt{10 - x^{2}}$ to the right side of the inequality.

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Step 2.1.1

Subtract $6$ from both sides of the inequality.

$\sqrt{10 - x^{2}} < 10 - x^{2} - 6$

Step 2.1.2

Subtract $6$ from $10$ .

$\sqrt{10 - x^{2}} < - x^{2} + 4$

Step 2.2

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{10 - x^{2}}}^{2} < {(- x^{2} + 4)}^{2}$

Step 2.3

Simplify each side of the inequality.

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Step 2.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10 - x^{2}}$ as ${(10 - x^{2})}^{\frac{1}{2}}$ .

${({(10 - x^{2})}^{\frac{1}{2}})}^{2} < {(- x^{2} + 4)}^{2}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Simplify ${({(10 - x^{2})}^{\frac{1}{2}})}^{2}$ .

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Step 2.3.2.1.1

Multiply the exponents in ${({(10 - x^{2})}^{\frac{1}{2}})}^{2}$ .

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Step 2.3.2.1.1.1

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

${(10 - x^{2})}^{\frac{1}{2} \cdot 2} < {(- x^{2} + 4)}^{2}$

Step 2.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.3.2.1.1.2.1

Cancel the common factor.

${(10 - x^{2})}^{\frac{1}{2} \cdot 2} < {(- x^{2} + 4)}^{2}$

Step 2.3.2.1.1.2.2

Rewrite the expression.

${(10 - x^{2})}^{1} < {(- x^{2} + 4)}^{2}$

Step 2.3.2.1.2

Simplify.

$10 - x^{2} < {(- x^{2} + 4)}^{2}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Simplify ${(- x^{2} + 4)}^{2}$ .

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Step 2.3.3.1.1

Rewrite ${(- x^{2} + 4)}^{2}$ as $(- x^{2} + 4) (- x^{2} + 4)$ .

$10 - x^{2} < (- x^{2} + 4) (- x^{2} + 4)$

Step 2.3.3.1.2

Expand $(- x^{2} + 4) (- x^{2} + 4)$ using the FOIL Method.

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Step 2.3.3.1.2.1

Apply the distributive property.

$10 - x^{2} < - x^{2} (- x^{2} + 4) + 4 (- x^{2} + 4)$

Step 2.3.3.1.2.2

Apply the distributive property.

$10 - x^{2} < - x^{2} (- x^{2}) - x^{2} \cdot 4 + 4 (- x^{2} + 4)$

Step 2.3.3.1.2.3

Apply the distributive property.

$10 - x^{2} < - x^{2} (- x^{2}) - x^{2} \cdot 4 + 4 (- x^{2}) + 4 \cdot 4$

Step 2.3.3.1.3

Simplify and combine like terms.

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Step 2.3.3.1.3.1

Simplify each term.

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Step 2.3.3.1.3.1.1

Rewrite using the commutative property of multiplication.

$10 - x^{2} < - 1 \cdot - 1 x^{2} x^{2} - x^{2} \cdot 4 + 4 (- x^{2}) + 4 \cdot 4$

Step 2.3.3.1.3.1.2

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Step 2.3.3.1.3.1.2.1

Move $x^{2}$ .

$10 - x^{2} < - 1 \cdot - 1 (x^{2} x^{2}) - x^{2} \cdot 4 + 4 (- x^{2}) + 4 \cdot 4$

Step 2.3.3.1.3.1.2.2

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

$10 - x^{2} < - 1 \cdot - 1 x^{2 + 2} - x^{2} \cdot 4 + 4 (- x^{2}) + 4 \cdot 4$

Step 2.3.3.1.3.1.2.3

Add $2$ and $2$ .

$10 - x^{2} < - 1 \cdot - 1 x^{4} - x^{2} \cdot 4 + 4 (- x^{2}) + 4 \cdot 4$

Step 2.3.3.1.3.1.3

Multiply $- 1$ by $- 1$ .

$10 - x^{2} < 1 x^{4} - x^{2} \cdot 4 + 4 (- x^{2}) + 4 \cdot 4$

Step 2.3.3.1.3.1.4

Multiply $x^{4}$ by $1$ .

$10 - x^{2} < x^{4} - x^{2} \cdot 4 + 4 (- x^{2}) + 4 \cdot 4$

Step 2.3.3.1.3.1.5

Multiply $4$ by $- 1$ .

$10 - x^{2} < x^{4} - 4 x^{2} + 4 (- x^{2}) + 4 \cdot 4$

Step 2.3.3.1.3.1.6

Multiply $- 1$ by $4$ .

$10 - x^{2} < x^{4} - 4 x^{2} - 4 x^{2} + 4 \cdot 4$

Step 2.3.3.1.3.1.7

Multiply $4$ by $4$ .

$10 - x^{2} < x^{4} - 4 x^{2} - 4 x^{2} + 16$

Step 2.3.3.1.3.2

Subtract $4 x^{2}$ from $- 4 x^{2}$ .

$10 - x^{2} < x^{4} - 8 x^{2} + 16$

Step 2.4

Solve for $x$ .

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Step 2.4.1

Rewrite so $x$ is on the left side of the inequality.

$x^{4} - 8 x^{2} + 16 > 10 - x^{2}$

Step 2.4.2

Move all terms containing $x$ to the left side of the inequality.

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Step 2.4.2.1

Add $x^{2}$ to both sides of the inequality.

$x^{4} - 8 x^{2} + 16 + x^{2} > 10$

Step 2.4.2.2

Add $- 8 x^{2}$ and $x^{2}$ .

$x^{4} - 7 x^{2} + 16 > 10$

Step 2.4.3

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} - 7 u + 16 = 10$

$u = x^{2}$

Step 2.4.4

Subtract $10$ from both sides of the equation.

$u^{2} - 7 u + 16 - 10 = 0$

Step 2.4.5

Subtract $10$ from $16$ .

$u^{2} - 7 u + 6 = 0$

Step 2.4.6

Factor $u^{2} - 7 u + 6$ using the AC method.

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Step 2.4.6.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 7$ .

$- 6, - 1$

Step 2.4.6.2

Write the factored form using these integers.

$(u - 6) (u - 1) = 0$

Step 2.4.7

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

$u - 6 = 0$

$u - 1 = 0$

Step 2.4.8

Set $u - 6$ equal to $0$ and solve for $u$ .

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Step 2.4.8.1

Set $u - 6$ equal to $0$ .

$u - 6 = 0$

Step 2.4.8.2

Add $6$ to both sides of the equation.

$u = 6$

Step 2.4.9

Set $u - 1$ equal to $0$ and solve for $u$ .

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Step 2.4.9.1

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Step 2.4.9.2

Add $1$ to both sides of the equation.

$u = 1$

Step 2.4.10

The final solution is all the values that make $(u - 6) (u - 1) = 0$ true.

$u = 6, 1$

Step 2.4.11

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = 6$

${(x^{2})}^{1} = 1$

Step 2.4.12

Solve the first equation for $x$ .

$x^{2} = 6$

Step 2.4.13

Solve the equation for $x$ .

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Step 2.4.13.1

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

$x = \pm \sqrt{6}$

Step 2.4.13.2

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

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Step 2.4.13.2.1

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

$x = \sqrt{6}$

Step 2.4.13.2.2

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

$x = - \sqrt{6}$

Step 2.4.13.2.3

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

$x = \sqrt{6}, - \sqrt{6}$

Step 2.4.14

Solve the second equation for $x$ .

${(x^{2})}^{1} = 1$

Step 2.4.15

Solve the equation for $x$ .

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Step 2.4.15.1

Remove parentheses.

$x^{2} = 1$

Step 2.4.15.2

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

$x = \pm \sqrt{1}$

Step 2.4.15.3

Any root of $1$ is $1$ .

$x = \pm 1$

Step 2.4.15.4

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

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Step 2.4.15.4.1

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

$x = 1$

Step 2.4.15.4.2

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

$x = - 1$

Step 2.4.15.4.3

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

$x = 1, - 1$

Step 2.4.16

The solution to $x^{4} - 7 x^{2} + 16 = 10$ is $x = \sqrt{6}, - \sqrt{6}, 1, - 1$ .

$x = \sqrt{6}, - \sqrt{6}, 1, - 1$

Step 2.5

Find the domain of $\sqrt{10 - x^{2}} - 4 + x^{2}$ .

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Step 2.5.1

Set the radicand in $\sqrt{10 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$10 - x^{2} \geq 0$

Step 2.5.2

Solve for $x$ .

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Step 2.5.2.1

Subtract $10$ from both sides of the inequality.

$- x^{2} \geq - 10$

Step 2.5.2.2

Divide each term in $- x^{2} \geq - 10$ by $- 1$ and simplify.

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Step 2.5.2.2.1

Divide each term in $- x^{2} \geq - 10$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 10}{- 1}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 10}{- 1}$

Step 2.5.2.2.2.2

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

$x^{2} \leq \frac{- 10}{- 1}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Divide $- 10$ by $- 1$ .

$x^{2} \leq 10$

Step 2.5.2.3

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

$\sqrt{x^{2}} \leq \sqrt{10}$

Step 2.5.2.4

Simplify the left side.

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Step 2.5.2.4.1

Pull terms out from under the radical.

$| x | \leq \sqrt{10}$

Step 2.5.2.5

Write $| x | \leq \sqrt{10}$ as a piecewise.

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Step 2.5.2.5.1

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

$x \geq 0$

Step 2.5.2.5.2

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

$x \leq \sqrt{10}$

Step 2.5.2.5.3

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

$x < 0$

Step 2.5.2.5.4

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

$- x \leq \sqrt{10}$

Step 2.5.2.5.5

Write as a piecewise.

${\begin{cases} x \leq \sqrt{10} & x \geq 0 \\ - x \leq \sqrt{10} & x < 0 \end{cases}$

Step 2.5.2.6

Find the intersection of $x \leq \sqrt{10}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{10}$

Step 2.5.2.7

Solve $- x \leq \sqrt{10}$ when $x < 0$ .

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Step 2.5.2.7.1

Divide each term in $- x \leq \sqrt{10}$ by $- 1$ and simplify.

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Step 2.5.2.7.1.1

Divide each term in $- x \leq \sqrt{10}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{10}}{- 1}$

Step 2.5.2.7.1.2

Simplify the left side.

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Step 2.5.2.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{10}}{- 1}$

Step 2.5.2.7.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{10}}{- 1}$

Step 2.5.2.7.1.3

Simplify the right side.

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Step 2.5.2.7.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{10}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{10}$

Step 2.5.2.7.1.3.2

Rewrite $- 1 \cdot \sqrt{10}$ as $- \sqrt{10}$ .

$x \geq - \sqrt{10}$

Step 2.5.2.7.2

Find the intersection of $x \geq - \sqrt{10}$ and $x < 0$ .

$- \sqrt{10} \leq x < 0$

Step 2.5.2.8

Find the union of the solutions.

$- \sqrt{10} \leq x \leq \sqrt{10}$

Step 2.5.3

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

$[- \sqrt{10}, \sqrt{10}]$

Step 2.6

Use each root to create test intervals.

$x < - \sqrt{10}$

$- \sqrt{10} < x < - \sqrt{6}$

$- \sqrt{6} < x < - 1$

$- 1 < x < 1$

$1 < x < \sqrt{6}$

$\sqrt{6} < x < \sqrt{10}$

$x > \sqrt{10}$

Step 2.7

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 2.7.1

Test a value on the interval $x < - \sqrt{10}$ to see if it makes the inequality true.

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Step 2.7.1.1

Choose a value on the interval $x < - \sqrt{10}$ and see if this value makes the original inequality true.

$x = - 6$

Step 2.7.1.2

Replace $x$ with $- 6$ in the original inequality.

$\sqrt{10 - {(- 6)}^{2}} + 6 < 10 - {(- 6)}^{2}$

Step 2.7.1.3

The left side is not equal to the right side, which means that the given statement is false.

False

Step 2.7.2

Test a value on the interval $- \sqrt{10} < x < - \sqrt{6}$ to see if it makes the inequality true.

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Step 2.7.2.1

Choose a value on the interval $- \sqrt{10} < x < - \sqrt{6}$ and see if this value makes the original inequality true.

$x = - 3$

Step 2.7.2.2

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

$\sqrt{10 - {(- 3)}^{2}} + 6 < 10 - {(- 3)}^{2}$

Step 2.7.2.3

The left side $7$ is not less than the right side $1$ , which means that the given statement is false.

False

Step 2.7.3

Test a value on the interval $- \sqrt{6} < x < - 1$ to see if it makes the inequality true.

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Step 2.7.3.1

Choose a value on the interval $- \sqrt{6} < x < - 1$ and see if this value makes the original inequality true.

$x = - 2$

Step 2.7.3.2

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

$\sqrt{10 - {(- 2)}^{2}} + 6 < 10 - {(- 2)}^{2}$

Step 2.7.3.3

The left side $8.44948974$ is not less than the right side $6$ , which means that the given statement is false.

False

Step 2.7.4

Test a value on the interval $- 1 < x < 1$ to see if it makes the inequality true.

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Step 2.7.4.1

Choose a value on the interval $- 1 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 2.7.4.2

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

$\sqrt{10 - {(0)}^{2}} + 6 < 10 - {(0)}^{2}$

Step 2.7.4.3

The left side $9.16227766$ is less than the right side $10$ , which means that the given statement is always true.

True

Step 2.7.5

Test a value on the interval $1 < x < \sqrt{6}$ to see if it makes the inequality true.

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Step 2.7.5.1

Choose a value on the interval $1 < x < \sqrt{6}$ and see if this value makes the original inequality true.

$x = 2$

Step 2.7.5.2

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

$\sqrt{10 - {(2)}^{2}} + 6 < 10 - {(2)}^{2}$

Step 2.7.5.3

The left side $8.44948974$ is not less than the right side $6$ , which means that the given statement is false.

False

Step 2.7.6

Test a value on the interval $\sqrt{6} < x < \sqrt{10}$ to see if it makes the inequality true.

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Step 2.7.6.1

Choose a value on the interval $\sqrt{6} < x < \sqrt{10}$ and see if this value makes the original inequality true.

$x = 3$

Step 2.7.6.2

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

$\sqrt{10 - {(3)}^{2}} + 6 < 10 - {(3)}^{2}$

Step 2.7.6.3

The left side $7$ is not less than the right side $1$ , which means that the given statement is false.

False

Step 2.7.7

Test a value on the interval $x > \sqrt{10}$ to see if it makes the inequality true.

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Step 2.7.7.1

Choose a value on the interval $x > \sqrt{10}$ and see if this value makes the original inequality true.

$x = 6$

Step 2.7.7.2

Replace $x$ with $6$ in the original inequality.

$\sqrt{10 - {(6)}^{2}} + 6 < 10 - {(6)}^{2}$

Step 2.7.7.3

The left side is not equal to the right side, which means that the given statement is false.

False

Step 2.7.8

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

$x < - \sqrt{10}$ False

$- \sqrt{10} < x < - \sqrt{6}$ False

$- \sqrt{6} < x < - 1$ False

$- 1 < x < 1$ True

$1 < x < \sqrt{6}$ False

$\sqrt{6} < x < \sqrt{10}$ False

$x > \sqrt{10}$ False

$x < - \sqrt{10}$ False

$- \sqrt{10} < x < - \sqrt{6}$ False

$- \sqrt{6} < x < - 1$ False

$- 1 < x < 1$ True

$1 < x < \sqrt{6}$ False

$\sqrt{6} < x < \sqrt{10}$ False

$x > \sqrt{10}$ False

Step 2.8

The solution consists of all of the true intervals.

$- 1 < x < 1$

Step 3

Find the union of the solutions.

$- 1 < x < 1$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$- 1 < x < 1$

Interval Notation:

$(- 1, 1)$

Step 5