Solve the Inequality for c square root of c+9- square root of c> square root of 3

$\sqrt{c + 9} - \sqrt{c} > \sqrt{3}$

Step 1

Add $\sqrt{c}$ to both sides of the inequality.

$\sqrt{c + 9} > \sqrt{3} + \sqrt{c}$

Step 2

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

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

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{c + 9}$ as ${(c + 9)}^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${({(c + 9)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(c + 9)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(\sqrt{3} + \sqrt{c})}^{2}$ .

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

$c + 9 > (\sqrt{3} + \sqrt{c}) (\sqrt{3} + \sqrt{c})$

Expand $(\sqrt{3} + \sqrt{c}) (\sqrt{3} + \sqrt{c})$ using the FOIL Method.

Apply the distributive property.

$c + 9 > \sqrt{3} (\sqrt{3} + \sqrt{c}) + \sqrt{c} (\sqrt{3} + \sqrt{c})$

Apply the distributive property.

$c + 9 > \sqrt{3} \sqrt{3} + \sqrt{3} \sqrt{c} + \sqrt{c} (\sqrt{3} + \sqrt{c})$

Apply the distributive property.

$c + 9 > \sqrt{3} \sqrt{3} + \sqrt{3} \sqrt{c} + \sqrt{c} \sqrt{3} + \sqrt{c} \sqrt{c}$

Simplify and combine like terms.

Simplify each term.

Combine using the product rule for radicals.

$c + 9 > \sqrt{3 \cdot 3} + \sqrt{3} \sqrt{c} + \sqrt{c} \sqrt{3} + \sqrt{c} \sqrt{c}$

Multiply $3$ by $3$ .

$c + 9 > \sqrt{9} + \sqrt{3} \sqrt{c} + \sqrt{c} \sqrt{3} + \sqrt{c} \sqrt{c}$

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

$c + 9 > \sqrt{3^{2}} + \sqrt{3} \sqrt{c} + \sqrt{c} \sqrt{3} + \sqrt{c} \sqrt{c}$

Pull terms out from under the radical, assuming positive real numbers.

$c + 9 > 3 + \sqrt{3} \sqrt{c} + \sqrt{c} \sqrt{3} + \sqrt{c} \sqrt{c}$

Combine using the product rule for radicals.

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c} \sqrt{3} + \sqrt{c} \sqrt{c}$

Combine using the product rule for radicals.

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c \cdot 3} + \sqrt{c} \sqrt{c}$

Multiply $\sqrt{c} \sqrt{c}$ .

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

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c \cdot 3} + {\sqrt{c}}^{1} \sqrt{c}$

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

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c \cdot 3} + {\sqrt{c}}^{1} {\sqrt{c}}^{1}$

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

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c \cdot 3} + {\sqrt{c}}^{1 + 1}$

Add $1$ and $1$ .

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c \cdot 3} + {\sqrt{c}}^{2}$

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

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

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

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

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c \cdot 3} + c^{\frac{1}{2} \cdot 2}$

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

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c \cdot 3} + c^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c \cdot 3} + c^{\frac{2}{2}}$

Rewrite the expression.

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c \cdot 3} + c^{1}$

Simplify.

$c + 9 > 3 + \sqrt{3 c} + \sqrt{c \cdot 3} + c$

Add $\sqrt{3 c}$ and $\sqrt{c \cdot 3}$ .

Reorder $c$ and $3$ .

$c + 9 > 3 + \sqrt{3 c} + \sqrt{3 \cdot c} + c$

Add $\sqrt{3 c}$ and $\sqrt{3 \cdot c}$ .

$c + 9 > 3 + 2 \sqrt{3 c} + c$

Step 4

Solve for $2 \sqrt{3 c}$ .

Rewrite so $2 \sqrt{3 c}$ is on the left side of the inequality.

$3 + 2 \sqrt{3 c} + c < c + 9$

Move all terms not containing $2 \sqrt{3 c}$ to the right side of the inequality.

Subtract $3$ from both sides of the inequality.

$2 \sqrt{3 c} + c < c + 9 - 3$

Subtract $c$ from both sides of the inequality.

$2 \sqrt{3 c} < c + 9 - 3 - c$

Combine the opposite terms in $c + 9 - 3 - c$ .

Subtract $c$ from $c$ .

$2 \sqrt{3 c} < 0 + 9 - 3$

Add $0$ and $9$ .

$2 \sqrt{3 c} < 9 - 3$

Subtract $3$ from $9$ .

$2 \sqrt{3 c} < 6$

Step 5

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

${(2 \sqrt{3 c})}^{2} < 6^{2}$

Step 6

Simplify each side of the inequality.

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

${(2 {(3 c)}^{\frac{1}{2}})}^{2} < 6^{2}$

Simplify the left side.

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

Apply the product rule to $3 c$ .

${(2 (3^{\frac{1}{2}} c^{\frac{1}{2}}))}^{2} < 6^{2}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $2 \cdot 3^{\frac{1}{2}} c^{\frac{1}{2}}$ .

${(2 \cdot 3^{\frac{1}{2}})}^{2} {(c^{\frac{1}{2}})}^{2} < 6^{2}$

Apply the product rule to $2 \cdot 3^{\frac{1}{2}}$ .

$2^{2} \cdot {(3^{\frac{1}{2}})}^{2} {(c^{\frac{1}{2}})}^{2} < 6^{2}$

Raise $2$ to the power of $2$ .

$4 \cdot {(3^{\frac{1}{2}})}^{2} {(c^{\frac{1}{2}})}^{2} < 6^{2}$

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

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

$4 \cdot 3^{\frac{1}{2} \cdot 2} {(c^{\frac{1}{2}})}^{2} < 6^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 \cdot 3^{\frac{1}{2} \cdot 2} {(c^{\frac{1}{2}})}^{2} < 6^{2}$

Rewrite the expression.

$4 \cdot 3^{1} {(c^{\frac{1}{2}})}^{2} < 6^{2}$

Evaluate the exponent.

$4 \cdot 3 {(c^{\frac{1}{2}})}^{2} < 6^{2}$

Multiply $4$ by $3$ .

$12 {(c^{\frac{1}{2}})}^{2} < 6^{2}$

Multiply the exponents in ${(c^{\frac{1}{2}})}^{2}$ .

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

$12 c^{\frac{1}{2} \cdot 2} < 6^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$12 c^{\frac{1}{2} \cdot 2} < 6^{2}$

Rewrite the expression.

$12 c^{1} < 6^{2}$

Simplify.

$12 c < 6^{2}$

Simplify the right side.

Raise $6$ to the power of $2$ .

$12 c < 36$

Step 7

Divide each term in $12 c < 36$ by $12$ and simplify.

Divide each term in $12 c < 36$ by $12$ .

$\frac{12 c}{12} < \frac{36}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 c}{12} < \frac{36}{12}$

Divide $c$ by $1$ .

$c < \frac{36}{12}$

Simplify the right side.

Divide $36$ by $12$ .

$c < 3$

Step 8

Find the domain of $\sqrt{c + 9} - \sqrt{c} - \sqrt{3}$ .

Set the radicand in $\sqrt{c + 9}$ greater than or equal to $0$ to find where the expression is defined.

$c + 9 \geq 0$

Subtract $9$ from both sides of the inequality.

$c \geq - 9$

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

$c \geq 0$

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

$[0, \infty)$

Step 9

Use each root to create test intervals.

$c < 0$

$0 < c < 3$

$c > 3$

Step 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 $c < 0$ to see if it makes the inequality true.

Choose a value on the interval $c < 0$ and see if this value makes the original inequality true.

$c = - 2$

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

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

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 $0 < c < 3$ to see if it makes the inequality true.

Choose a value on the interval $0 < c < 3$ and see if this value makes the original inequality true.

$c = 2$

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

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

The left side $1.90241122$ is greater than the right side $1.7320508$ , which means that the given statement is always true.

True

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

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

$c = 6$

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

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

The left side $1.4234936$ is not greater than the right side $1.7320508$ , which means that the given statement is false.

False

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

$c < 0$ False

$0 < c < 3$ True

$c > 3$ False

$c < 0$ False

$0 < c < 3$ True

$c > 3$ False

Step 11

The solution consists of all of the true intervals.

$0 < c < 3$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$0 < c < 3$

Interval Notation:

$(0, 3)$

Step 13

Solve the Inequality for c square root of c+9- square root of c> square root of 3

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