Graph square root of c+28- square root of c<=2

$\sqrt{c + 28} - \sqrt{c} \leq 2$

Step 1

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

$\sqrt{c + 28} \leq 2 + \sqrt{c}$

Step 2

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

${\sqrt{c + 28}}^{2} \leq {(2 + \sqrt{c})}^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{c + 28}$ as ${(c + 28)}^{\frac{1}{2}}$ .

${({(c + 28)}^{\frac{1}{2}})}^{2} \leq {(2 + \sqrt{c})}^{2}$

Simplify the left side.

Simplify ${({(c + 28)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(c + 28)}^{\frac{1}{2}})}^{2}$ .

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

${(c + 28)}^{\frac{1}{2} \cdot 2} \leq {(2 + \sqrt{c})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(c + 28)}^{\frac{1}{2} \cdot 2} \leq {(2 + \sqrt{c})}^{2}$

Rewrite the expression.

${(c + 28)}^{1} \leq {(2 + \sqrt{c})}^{2}$

Simplify.

$c + 28 \leq {(2 + \sqrt{c})}^{2}$

Simplify the right side.

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

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

$c + 28 \leq (2 + \sqrt{c}) (2 + \sqrt{c})$

Expand $(2 + \sqrt{c}) (2 + \sqrt{c})$ using the FOIL Method.

Apply the distributive property.

$c + 28 \leq 2 (2 + \sqrt{c}) + \sqrt{c} (2 + \sqrt{c})$

Apply the distributive property.

$c + 28 \leq 2 \cdot 2 + 2 \sqrt{c} + \sqrt{c} (2 + \sqrt{c})$

Apply the distributive property.

$c + 28 \leq 2 \cdot 2 + 2 \sqrt{c} + \sqrt{c} \cdot 2 + \sqrt{c} \sqrt{c}$

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $2$ .

$c + 28 \leq 4 + 2 \sqrt{c} + \sqrt{c} \cdot 2 + \sqrt{c} \sqrt{c}$

Move $2$ to the left of $\sqrt{c}$ .

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \cdot \sqrt{c} + \sqrt{c} \sqrt{c}$

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

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

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \sqrt{c} + {\sqrt{c}}^{1} \sqrt{c}$

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

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \sqrt{c} + {\sqrt{c}}^{1} {\sqrt{c}}^{1}$

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

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \sqrt{c} + {\sqrt{c}}^{1 + 1}$

Add $1$ and $1$ .

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \sqrt{c} + {\sqrt{c}}^{2}$

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

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

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \sqrt{c} + {(c^{\frac{1}{2}})}^{2}$

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

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \sqrt{c} + c^{\frac{1}{2} \cdot 2}$

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

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \sqrt{c} + c^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \sqrt{c} + c^{\frac{2}{2}}$

Rewrite the expression.

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \sqrt{c} + c^{1}$

Simplify.

$c + 28 \leq 4 + 2 \sqrt{c} + 2 \sqrt{c} + c$

Add $2 \sqrt{c}$ and $2 \sqrt{c}$ .

$c + 28 \leq 4 + 4 \sqrt{c} + c$

Step 4

Solve for $4 \sqrt{c}$ .

Rewrite so $4 \sqrt{c}$ is on the left side of the inequality.

$4 + 4 \sqrt{c} + c \geq c + 28$

Move all terms not containing $4 \sqrt{c}$ to the right side of the inequality.

Subtract $4$ from both sides of the inequality.

$4 \sqrt{c} + c \geq c + 28 - 4$

Subtract $c$ from both sides of the inequality.

$4 \sqrt{c} \geq c + 28 - 4 - c$

Combine the opposite terms in $c + 28 - 4 - c$ .

Subtract $c$ from $c$ .

$4 \sqrt{c} \geq 0 + 28 - 4$

Add $0$ and $28$ .

$4 \sqrt{c} \geq 28 - 4$

Subtract $4$ from $28$ .

$4 \sqrt{c} \geq 24$

Step 5

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

${(4 \sqrt{c})}^{2} \geq 24^{2}$

Step 6

Simplify each side of the inequality.

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

${(4 c^{\frac{1}{2}})}^{2} \geq 24^{2}$

Simplify the left side.

Simplify ${(4 c^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $4 c^{\frac{1}{2}}$ .

$4^{2} {(c^{\frac{1}{2}})}^{2} \geq 24^{2}$

Raise $4$ to the power of $2$ .

$16 {(c^{\frac{1}{2}})}^{2} \geq 24^{2}$

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

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

$16 c^{\frac{1}{2} \cdot 2} \geq 24^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$16 c^{\frac{1}{2} \cdot 2} \geq 24^{2}$

Rewrite the expression.

$16 c^{1} \geq 24^{2}$

Simplify.

$16 c \geq 24^{2}$

Simplify the right side.

Raise $24$ to the power of $2$ .

$16 c \geq 576$

Step 7

Divide each term in $16 c \geq 576$ by $16$ and simplify.

Divide each term in $16 c \geq 576$ by $16$ .

$\frac{16 c}{16} \geq \frac{576}{16}$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 c}{16} \geq \frac{576}{16}$

Divide $c$ by $1$ .

$c \geq \frac{576}{16}$

Simplify the right side.

Divide $576$ by $16$ .

$c \geq 36$

Step 8

Find the domain of $\sqrt{c + 28} - \sqrt{c} - 2$ .

Set the radicand in $\sqrt{c + 28}$ greater than or equal to $0$ to find where the expression is defined.

$c + 28 \geq 0$

Subtract $28$ from both sides of the inequality.

$c \geq - 28$

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

The solution consists of all of the true intervals.

$c \geq 36$

Step 10

Graph square root of c+28- square root of c<=2

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