Simplify ( square root of x+1)÷( square root of 3-x)

$\sqrt{x + 1} \div \sqrt{3 - x}$

Step 1

Rewrite the division as a fraction.

$\frac{\sqrt{x + 1}}{\sqrt{3 - x}}$

Step 2

Multiply $\frac{\sqrt{x + 1}}{\sqrt{3 - x}}$ by $\frac{\sqrt{3 - x}}{\sqrt{3 - x}}$ .

$\frac{\sqrt{x + 1}}{\sqrt{3 - x}} \cdot \frac{\sqrt{3 - x}}{\sqrt{3 - x}}$

Step 3

Combine and simplify the denominator.

Multiply $\frac{\sqrt{x + 1}}{\sqrt{3 - x}}$ by $\frac{\sqrt{3 - x}}{\sqrt{3 - x}}$ .

$\frac{\sqrt{x + 1} \sqrt{3 - x}}{\sqrt{3 - x} \sqrt{3 - x}}$

Raise $\sqrt{3 - x}$ to the power of $1$ .

$\frac{\sqrt{x + 1} \sqrt{3 - x}}{{\sqrt{3 - x}}^{1} \sqrt{3 - x}}$

Raise $\sqrt{3 - x}$ to the power of $1$ .

$\frac{\sqrt{x + 1} \sqrt{3 - x}}{{\sqrt{3 - x}}^{1} {\sqrt{3 - x}}^{1}}$

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

$\frac{\sqrt{x + 1} \sqrt{3 - x}}{{\sqrt{3 - x}}^{1 + 1}}$

Add $1$ and $1$ .

$\frac{\sqrt{x + 1} \sqrt{3 - x}}{{\sqrt{3 - x}}^{2}}$

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$\frac{\sqrt{x + 1} \sqrt{3 - x}}{3 - x}$

Step 4

Combine using the product rule for radicals.

$\frac{\sqrt{(x + 1) (3 - x)}}{3 - x}$