Solve the Inequality for x x+2> square root of 10-x^2

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

Step 1

Since the radical is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Step 2

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

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

Step 3

Simplify each side of the inequality.

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

Expand $(x + 2) (x + 2)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $2$ to the left of $x$ .

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

Multiply $2$ by $2$ .

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

Add $2 x$ and $2 x$ .

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

Step 4

Solve for $x$ .

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

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

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

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

$x^{2} + 4 x + 4 + x^{2} > 10$

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

$2 x^{2} + 4 x + 4 > 10$

Convert the inequality to an equation.

$2 x^{2} + 4 x + 4 = 10$

Subtract $10$ from both sides of the equation.

$2 x^{2} + 4 x + 4 - 10 = 0$

Subtract $10$ from $4$ .

$2 x^{2} + 4 x - 6 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 4 x - 6$ .

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

$2 (x^{2}) + 4 x - 6 = 0$

Factor $2$ out of $4 x$ .

$2 (x^{2}) + 2 (2 x) - 6 = 0$

Factor $2$ out of $- 6$ .

$2 x^{2} + 2 (2 x) + 2 \cdot - 3 = 0$

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

$2 (x^{2} + 2 x) + 2 \cdot - 3 = 0$

Factor $2$ out of $2 (x^{2} + 2 x) + 2 \cdot - 3$ .

$2 (x^{2} + 2 x - 3) = 0$

Factor.

Factor $x^{2} + 2 x - 3$ 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 $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$2 ((x - 1) (x + 3)) = 0$

Remove unnecessary parentheses.

$2 (x - 1) (x + 3) = 0$

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

$x - 1 = 0$

$x + 3 = 0$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $2 (x - 1) (x + 3) = 0$ true.

$x = 1, - 3$

Step 5

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

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.

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

Step 6

Use each root to create test intervals.

$x < - \sqrt{10}$

$- \sqrt{10} < x < - 3$

$- 3 < x < 1$

$1 < x < \sqrt{10}$

$x > \sqrt{10}$

Step 7

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.

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

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 $- \sqrt{10} < x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $- \sqrt{10} < x < - 3$ and see if this value makes the original inequality true.

$x = - 3.08$

Replace $x$ with $- 3.08$ in the original inequality.

$(- 3.08) + 2 > \sqrt{10 - {(- 3.08)}^{2}}$

The left side $- 1.08$ is not greater than the right side $0.71665891$ , which means that the given statement is false.

False

Test a value on the interval $- 3 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $2$ is not greater than the right side $3.16227766$ , which means that the given statement is false.

False

Test a value on the interval $1 < x < \sqrt{10}$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < \sqrt{10}$ and see if this value makes the original inequality true.

$x = 2$

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

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

The left side $4$ is greater than the right side $2.44948974$ , which means that the given statement is always true.

True

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.

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

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

$- \sqrt{10} < x < - 3$ False

$- 3 < x < 1$ False

$1 < x < \sqrt{10}$ True

$x > \sqrt{10}$ False

$x < - \sqrt{10}$ False

$- \sqrt{10} < x < - 3$ False

$- 3 < x < 1$ False

$1 < x < \sqrt{10}$ True

$x > \sqrt{10}$ False

Step 8

The solution consists of all of the true intervals.

$1 < x < \sqrt{10}$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$1 < x < \sqrt{10}$

Interval Notation:

$(1, \sqrt{10})$

Step 10

Solve the Inequality for x x+2> square root of 10-x^2

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