Algebra Examples

\sqrt{c + 9} - \sqrt{c} > \sqrt{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.

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Step 3.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt{c + 9}$ as

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

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify

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

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Step 3.2.1.1

Multiply the exponents in

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

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Step 3.2.1.1.1

Apply the power rule and multiply exponents,

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

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

Step 3.2.1.1.2

Cancel the common factor of

$2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.1.2

Simplify.

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

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify

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

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Step 3.3.1.1

Rewrite

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

$(\sqrt{3} + \sqrt{c}) (\sqrt{3} + \sqrt{c})$ .

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

Step 3.3.1.2

Expand

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

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Step 3.3.1.2.1

Apply the distributive property.

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

Step 3.3.1.2.2

Apply the distributive property.

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

Step 3.3.1.2.3

Apply the distributive property.

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

Step 3.3.1.3

Simplify and combine like terms.

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Step 3.3.1.3.1

Simplify each term.

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Step 3.3.1.3.1.1

Combine using the product rule for radicals.

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

Step 3.3.1.3.1.2

Multiply

$3$ by

$3$ .

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

Step 3.3.1.3.1.3

Rewrite

$9$ as

$3^{2}$ .

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

Step 3.3.1.3.1.4

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

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

Step 3.3.1.3.1.5

Combine using the product rule for radicals.

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

Step 3.3.1.3.1.6

Combine using the product rule for radicals.

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

Step 3.3.1.3.1.7

Multiply

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

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Step 3.3.1.3.1.7.1

Raise

$\sqrt{c}$ to the power of

$1$ .

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

Step 3.3.1.3.1.7.2

Raise

$\sqrt{c}$ to the power of

$1$ .

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

Step 3.3.1.3.1.7.3

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.3.1.3.1.7.4

Add

$1$ and

$1$ .

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

Step 3.3.1.3.1.8

Rewrite

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

$c$ .

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Step 3.3.1.3.1.8.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt{c}$ as

$c^{\frac{1}{2}}$ .

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

Step 3.3.1.3.1.8.2

Apply the power rule and multiply exponents,

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

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

Step 3.3.1.3.1.8.3

Combine

$\frac{1}{2}$ and

$2$ .

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

Step 3.3.1.3.1.8.4

Cancel the common factor of

$2$ .

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Step 3.3.1.3.1.8.4.1

Cancel the common factor.

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

Step 3.3.1.3.1.8.4.2

Rewrite the expression.

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

Step 3.3.1.3.1.8.5

Simplify.

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

Step 3.3.1.3.2

Add

$\sqrt{3 c}$ and

$\sqrt{c \cdot 3}$ .

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Step 3.3.1.3.2.1

Reorder

$c$ and

$3$ .

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

Step 3.3.1.3.2.2

Add

$\sqrt{3 c}$ and

$\sqrt{3 \cdot c}$ .

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

Step 4

Solve for

$2 \sqrt{3 c}$ .

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Step 4.1

Rewrite so

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

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

Step 4.2

Move all terms not containing

$2 \sqrt{3 c}$ to the right side of the inequality.

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Step 4.2.1

Subtract

$3$ from both sides of the inequality.

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

Step 4.2.2

Subtract

$c$ from both sides of the inequality.

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

Step 4.2.3

Combine the opposite terms in

$c + 9 - 3 - c$ .

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Step 4.2.3.1

Subtract

$c$ from

$c$ .

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

Step 4.2.3.2

Add

$0$ and

$9$ .

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

Step 4.2.4

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.

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Step 6.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt{3 c}$ as

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

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

Step 6.2

Simplify the left side.

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Step 6.2.1

Simplify

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

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Step 6.2.1.1

Apply the product rule to

$3 c$ .

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

Step 6.2.1.2

Use the power rule

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

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Step 6.2.1.2.1

Apply the product rule to

$2 \cdot 3^{\frac{1}{2}} c^{\frac{1}{2}}$ .

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

Step 6.2.1.2.2

Apply the product rule to

$2 \cdot 3^{\frac{1}{2}}$ .

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

Step 6.2.1.3

Raise

$2$ to the power of

$2$ .

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

Step 6.2.1.4

Multiply the exponents in

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

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Step 6.2.1.4.1

Apply the power rule and multiply exponents,

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

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

Step 6.2.1.4.2

Cancel the common factor of

$2$ .

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Step 6.2.1.4.2.1

Cancel the common factor.

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

Step 6.2.1.4.2.2

Rewrite the expression.

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

Step 6.2.1.5

Evaluate the exponent.

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

Step 6.2.1.6

Multiply

$4$ by

$3$ .

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

Step 6.2.1.7

Multiply the exponents in

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

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Step 6.2.1.7.1

Apply the power rule and multiply exponents,

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

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

Step 6.2.1.7.2

Cancel the common factor of

$2$ .

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Step 6.2.1.7.2.1

Cancel the common factor.

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

Step 6.2.1.7.2.2

Rewrite the expression.

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

Step 6.2.1.8

Simplify.

$12 c < 6^{2}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Raise

$6$ to the power of

$2$ .

$12 c < 36$

Step 7

Divide each term in

$12 c < 36$ by

$12$ and simplify.

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Step 7.1

Divide each term in

$12 c < 36$ by

$12$ .

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

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of

$12$ .

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Step 7.2.1.1

Cancel the common factor.

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

Step 7.2.1.2

Divide

$c$ by

$1$ .

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

Step 7.3

Simplify the right side.

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Step 7.3.1

Divide

$36$ by

$12$ .

$c < 3$

Step 8

Find the domain of

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

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Step 8.1

Set the radicand in

$\sqrt{c + 9}$ greater than or equal to

$0$ to find where the expression is defined.

$c + 9 \geq 0$

Step 8.2

Subtract

$9$ from both sides of the inequality.

$c \geq - 9$

Step 8.3

Set the radicand in

$\sqrt{c}$ greater than or equal to

$0$ to find where the expression is defined.

$c \geq 0$

Step 8.4

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.

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Step 10.1

Test a value on the interval

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

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Step 10.1.1

Choose a value on the interval

$c < 0$ and see if this value makes the original inequality true.

$c = - 2$

Step 10.1.2

Replace

$c$ with

$- 2$ in the original inequality.

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

Step 10.1.3

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

False

Step 10.2

Test a value on the interval

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

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Step 10.2.1

Choose a value on the interval

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

$c = 2$

Step 10.2.2

Replace

$c$ with

$2$ in the original inequality.

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

Step 10.2.3

The left side

$1.90241122$ is greater than the right side

$1.7320508$ , which means that the given statement is always true.

True

Step 10.3

Test a value on the interval

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

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Step 10.3.1

Choose a value on the interval

$c > 3$ and see if this value makes the original inequality true.

$c = 6$

Step 10.3.2

Replace

$c$ with

$6$ in the original inequality.

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

Step 10.3.3

The left side

$1.4234936$ is not greater than the right side

$1.7320508$ , which means that the given statement is false.

False

Step 10.4

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