Solve the Inequality for x square root of 10-x^2+6<|10-x^2|

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

Step 1

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

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

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

Solve the inequality.

Subtract $10$ from both sides of the inequality.

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

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $- 10$ by $- 1$ .

$x^{2} \leq 10$

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

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

Simplify the left side.

Pull terms out from under the radical.

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

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

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

$x \geq 0$

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

$x \leq \sqrt{10}$

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

$x < 0$

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

$- x \leq \sqrt{10}$

Write as a piecewise.

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

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

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

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

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

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

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

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

$x \geq - \sqrt{10}$

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

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

Find the union of the solutions.

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

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

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

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

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

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

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

Solve for $x$ .

Subtract $10$ from both sides of the inequality.

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

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $- 10$ by $- 1$ .

$x^{2} \leq 10$

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

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

Simplify the left side.

Pull terms out from under the radical.

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

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

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

$x \geq 0$

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

$x \leq \sqrt{10}$

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

$x < 0$

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

$- x \leq \sqrt{10}$

Write as a piecewise.

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

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

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

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

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

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

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

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

$x \geq - \sqrt{10}$

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

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

Find the union of the solutions.

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

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

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

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

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

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

$10 - x^{2} < 0$

Solve the inequality.

Subtract $10$ from both sides of the inequality.

$- x^{2} < - 10$

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $- 10$ by $- 1$ .

$x^{2} > 10$

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

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

Simplify the left side.

Pull terms out from under the radical.

$| x | > \sqrt{10}$

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

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

$x \geq 0$

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

$x > \sqrt{10}$

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

$x < 0$

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

$- x > \sqrt{10}$

Write as a piecewise.

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

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

$x > \sqrt{10}$

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

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

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

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

$x < - \sqrt{10}$

Find the union of the solutions.

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

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

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

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

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

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

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

Solve for $x$ .

Subtract $10$ from both sides of the inequality.

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

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $- 10$ by $- 1$ .

$x^{2} \leq 10$

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

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

Simplify the left side.

Pull terms out from under the radical.

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

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

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

$x \geq 0$

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

$x \leq \sqrt{10}$

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

$x < 0$

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

$- x \leq \sqrt{10}$

Write as a piecewise.

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

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

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

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

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

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

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

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

$x \geq - \sqrt{10}$

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

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

Find the union of the solutions.

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

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

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

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

$No solution$

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 $x$ .

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

Subtract $6$ from both sides of the inequality.

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

Subtract $6$ from $10$ .

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

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

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

Simplify each side of the inequality.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

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

Move $x^{2}$ .

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

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

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

Add $2$ and $2$ .

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

Multiply $- 1$ by $- 1$ .

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

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

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

Multiply $4$ by $- 1$ .

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

Multiply $- 1$ by $4$ .

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

Multiply $4$ by $4$ .

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

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

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

Solve for $x$ .

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

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

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

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

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

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

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

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

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

$u = x^{2}$

Subtract $10$ from both sides of the equation.

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

Subtract $10$ from $16$ .

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

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

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 $- 7$ .

$- 6, - 1$

Write the factored form using these integers.

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

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

$u - 6 = 0$

$u - 1 = 0$

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

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

$u - 6 = 0$

Add $6$ to both sides of the equation.

$u = 6$

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

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

$u - 1 = 0$

Add $1$ to both sides of the equation.

$u = 1$

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

$u = 6, 1$

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

$x^{2} = 6$

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

Solve the first equation for $x$ .

$x^{2} = 6$

Solve the equation for $x$ .

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

$x = \pm \sqrt{6}$

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

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

$x = \sqrt{6}$

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

$x = - \sqrt{6}$

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

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

Solve the second equation for $x$ .

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

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 1$

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

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

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

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

$x = 1$

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

$x = - 1$

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

$x = 1, - 1$

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

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

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

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

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

Solve for $x$ .

Subtract $10$ from both sides of the inequality.

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

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $- 10$ by $- 1$ .

$x^{2} \leq 10$

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

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

Simplify the left side.

Pull terms out from under the radical.

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

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

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

$x \geq 0$

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

$x \leq \sqrt{10}$

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

$x < 0$

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

$- x \leq \sqrt{10}$

Write as a piecewise.

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

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

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

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

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

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.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

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

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

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

$x \geq - \sqrt{10}$

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

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

Find the union of the solutions.

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

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

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

Use each root to create test intervals.

$x < - \sqrt{10}$

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

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

$- 1 < x < 1$

$1 < x < \sqrt{6}$

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

$x > \sqrt{10}$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = - 6$

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

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

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

False

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

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

$x = - 3$

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

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

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

False

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

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

$x = - 2$

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

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

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

False

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

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

$x = 0$

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

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

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

True

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

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

$x = 2$

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

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

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

False

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

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

$x = 3$

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

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

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

False

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

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

$x = 6$

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

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

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

False

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

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

Solve the Inequality for x square root of 10-x^2+6<|10-x^2|

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