Solve the Inequality for x square root of x+3> square root of 9-x^2

$\sqrt{x + 3} > \sqrt{9 - x^{2}}$

Step 1

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

${\sqrt{x + 3}}^{2} > {\sqrt{9 - x^{2}}}^{2}$

Step 2

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 3}$ as ${(x + 3)}^{\frac{1}{2}}$ .

${({(x + 3)}^{\frac{1}{2}})}^{2} > {\sqrt{9 - x^{2}}}^{2}$

Simplify the left side.

Simplify ${({(x + 3)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x + 3)}^{\frac{1}{2}})}^{2}$ .

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

${(x + 3)}^{\frac{1}{2} \cdot 2} > {\sqrt{9 - x^{2}}}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x + 3)}^{\frac{1}{2} \cdot 2} > {\sqrt{9 - x^{2}}}^{2}$

Rewrite the expression.

${(x + 3)}^{1} > {\sqrt{9 - x^{2}}}^{2}$

Simplify.

$x + 3 > {\sqrt{9 - x^{2}}}^{2}$

Simplify the right side.

Simplify ${\sqrt{9 - x^{2}}}^{2}$ .

Rewrite $9$ as $3^{2}$ .

$x + 3 > {\sqrt{3^{2} - x^{2}}}^{2}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = x$ .

$x + 3 > {\sqrt{(3 + x) (3 - x)}}^{2}$

Rewrite ${\sqrt{(3 + x) (3 - x)}}^{2}$ as $(3 + x) (3 - x)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{(3 + x) (3 - x)}$ as ${((3 + x) (3 - x))}^{\frac{1}{2}}$ .

$x + 3 > {({((3 + x) (3 - x))}^{\frac{1}{2}})}^{2}$

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

$x + 3 > {((3 + x) (3 - x))}^{\frac{1}{2} \cdot 2}$

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

$x + 3 > {((3 + x) (3 - x))}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x + 3 > {((3 + x) (3 - x))}^{\frac{2}{2}}$

Rewrite the expression.

$x + 3 > {((3 + x) (3 - x))}^{1}$

Simplify.

$x + 3 > (3 + x) (3 - x)$

Expand $(3 + x) (3 - x)$ using the FOIL Method.

Apply the distributive property.

$x + 3 > 3 (3 - x) + x (3 - x)$

Apply the distributive property.

$x + 3 > 3 \cdot 3 + 3 (- x) + x (3 - x)$

Apply the distributive property.

$x + 3 > 3 \cdot 3 + 3 (- x) + x \cdot 3 + x (- x)$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$x + 3 > 9 + 3 (- x) + x \cdot 3 + x (- x)$

Multiply $- 1$ by $3$ .

$x + 3 > 9 - 3 x + x \cdot 3 + x (- x)$

Move $3$ to the left of $x$ .

$x + 3 > 9 - 3 x + 3 \cdot x + x (- x)$

Rewrite using the commutative property of multiplication.

$x + 3 > 9 - 3 x + 3 x - x \cdot x$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$x + 3 > 9 - 3 x + 3 x - (x \cdot x)$

Multiply $x$ by $x$ .

$x + 3 > 9 - 3 x + 3 x - x^{2}$

Add $- 3 x$ and $3 x$ .

$x + 3 > 9 + 0 - x^{2}$

Add $9$ and $0$ .

$x + 3 > 9 - x^{2}$

Step 3

Solve for $x$ .

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

$x + 3 + x^{2} > 9$

Convert the inequality to an equation.

$x + 3 + x^{2} = 9$

Subtract $9$ from both sides of the equation.

$x + 3 + x^{2} - 9 = 0$

Subtract $9$ from $3$ .

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

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

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

Factor $u + u^{2} - 6$ 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 $1$ .

$- 2, 3$

Write the factored form using these integers.

$(u - 2) (u + 3) = 0$

Replace all occurrences of $u$ with $x$ .

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

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

$x - 2 = 0$

$x + 3 = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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 $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

Step 4

Find the domain of $\sqrt{x + 3} - \sqrt{(3 + x) (3 - x)}$ .

Set the radicand in $\sqrt{x + 3}$ greater than or equal to $0$ to find where the expression is defined.

$x + 3 \geq 0$

Subtract $3$ from both sides of the inequality.

$x \geq - 3$

Set the radicand in $\sqrt{(3 + x) (3 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(3 + x) (3 - x) \geq 0$

Solve for $x$ .

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

$3 + x = 0$

$3 - x = 0$

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

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

$3 + x = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

$3 - x = 0$

Solve $3 - x = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

The final solution is all the values that make $(3 + x) (3 - x) \geq 0$ true.

$x = - 3, 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 3$

$x > 3$

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 $x < - 3$ to see if it makes the inequality true.

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

$x = - 6$

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

$(3 - 6) (3 - (- 6)) \geq 0$

The left side $- 27$ is less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$(3 + 0) (3 - (0)) \geq 0$

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 6$

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

$(3 + 6) (3 - (6)) \geq 0$

The left side $- 27$ is less than the right side $0$ , which means that the given statement is false.

False

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

$x < - 3$ False

$- 3 < x < 3$ True

$x > 3$ False

$x < - 3$ False

$- 3 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 3 \leq x \leq 3$

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

$[- 3, 3]$

Step 5

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 2$

$2 < x < 3$

$x > 3$

Step 6

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 $x < - 3$ to see if it makes the inequality true.

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

$x = - 6$

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

$\sqrt{(- 6) + 3} > \sqrt{9 - {(- 6)}^{2}}$

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 $- 3 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$\sqrt{(0) + 3} > \sqrt{9 - {(0)}^{2}}$

The left side $1.7320508$ is not greater than the right side $3$ , which means that the given statement is false.

False

Test a value on the interval $2 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Replace $x$ with $2.5$ in the original inequality.

$\sqrt{(2.5) + 3} > \sqrt{9 - {(2.5)}^{2}}$

The left side $2.34520787$ is greater than the right side $1.65831239$ , which means that the given statement is always true.

True

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

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

$x = 6$

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

$\sqrt{(6) + 3} > \sqrt{9 - {(6)}^{2}}$

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.

$x < - 3$ False

$- 3 < x < 2$ False

$2 < x < 3$ True

$x > 3$ False

$x < - 3$ False

$- 3 < x < 2$ False

$2 < x < 3$ True

$x > 3$ False

Step 7

The solution consists of all of the true intervals.

$2 < x < 3$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$2 < x < 3$

Interval Notation:

$(2, 3)$

Step 9

Solve the Inequality for x square root of x+3> square root of 9-x^2

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