Find the Domain ( square root of x+2)÷( square root of 5-x)

$\sqrt{x + 2} \div \sqrt{5 - x}$

Set the radicand in $\sqrt{x + 2}$ greater than or equal to $0$ to find where the expression is defined.

$x + 2 \geq 0$

Solve for $x$ .

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Subtract $2$ from both sides of the inequality.

$x \geq - 2$

The solution consists of all of the true intervals.

$x \geq - 2$

Set the radicand in $\sqrt{5 - x}$ greater than or equal to $0$ to find where the expression is defined.

$5 - x \geq 0$

Solve for $x$ .

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Subtract $5$ from both sides of the inequality.

$- x \geq - 5$

Multiply each term in $- x \geq - 5$ by $- 1$

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Multiply each term in $- x \geq - 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$(- x) \cdot - 1 \leq (- 5) \cdot - 1$

Multiply $(- x) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 x \leq (- 5) \cdot - 1$

Multiply $x$ by $1$ .

$x \leq (- 5) \cdot - 1$

Multiply $- 5$ by $- 1$ .

$x \leq 5$

The solution consists of all of the true intervals.

$x \leq 5$

Set the denominator in $\frac{\sqrt{x + 2}}{\sqrt{5 - x}}$ equal to $0$ to find where the expression is undefined.

$\sqrt{5 - x} = 0$

Solve for $x$ .

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

${\sqrt{5 - x}}^{2} = 0^{2}$

Simplify each side of the equation.

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Multiply the exponents in ${({(5 - x)}^{\frac{1}{2}})}^{2}$ .

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

${(5 - x)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

${(5 - x)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Rewrite the expression.

${(5 - x)}^{1} = 0^{2}$

Simplify.

$5 - x = 0^{2}$

Raising $0$ to any positive power yields $0$ .

$5 - x = 0$

Solve for $x$ .

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Subtract $5$ from both sides of the equation.

$- x = - 5$

Multiply each term in $- x = - 5$ by $- 1$

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Multiply each term in $- x = - 5$ by $- 1$ .

$(- x) \cdot - 1 = (- 5) \cdot - 1$

Multiply $(- x) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 x = (- 5) \cdot - 1$

Multiply $x$ by $1$ .

$x = (- 5) \cdot - 1$

Multiply $- 5$ by $- 1$ .

$x = 5$

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

Interval Notation:

$[- 2, 5)$

Set-Builder Notation:

${x | - 2 \leq x < 5}$

Find the Domain ( square root of x+2)÷( square root of 5-x)

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