Question

# To calculate: The simplified form of the expression displaystylefrac{sqrt{{12}}}{sqrt{{{x}+{1}}}}.

To calculate: The simplified form of the expression $$\displaystyle\frac{\sqrt{{12}}}{\sqrt{{{x}+{1}}}}.$$

2020-12-17

Formula used:
Product property of radicals: $$\displaystyle{\sqrt[{{n}}]{{a}}}{\sqrt[{{n}}]{{b}}}={\sqrt[{{n}}]{{{a}{b}}}}$$
Power property of radicals: $$\displaystyle{\left({a}^{m}\right)}^{n}={a}^{{{m}{n}}}$$
Calculation:
Consider the provided expression, $$\displaystyle\frac{\sqrt{{12}}}{\sqrt{{{x}+{1}}}}$$
Multiply numerator and denominator by $$\displaystyle\sqrt{{{x}+{1}}}$$ so that the radicand in the denominator is a perfect square.
Therefore,
$$\displaystyle\frac{\sqrt{{12}}}{\sqrt{{{x}+{1}}}}=\frac{\sqrt{{12}}}{\sqrt{{{x}+{1}}}}\cdot\frac{\sqrt{{{x}+{1}}}}{\sqrt{{{x}+{1}}}}$$
Here, $$\displaystyle{x}>-{1}$$
Apply the product property of radicals:
$$\displaystyle{\sqrt[{{n}}]{{a}}}{\sqrt[{{n}}]{{b}}}={\sqrt[{{n}}]{{{a}{b}}}}$$
Therefore,
$$\displaystyle\frac{\sqrt{{12}}}{\sqrt{{{x}+{1}}}}=\frac{{\sqrt{{{4}\cdot{3}}}\sqrt{{{x}+{1}}}}}{\sqrt{{{\left({x}+{1}\right)}{\left({x}+{1}\right)}}}}$$
$$\displaystyle=\frac{{{\sqrt[{{2}}]{{3}}}\sqrt{{{x}+{1}}}}}{\sqrt{{{\left({x}+{1}\right)}^{2}}}}$$
Apply power property $$\displaystyle{\left({a}^{m}\right)}^{n}={a}^{{{m}{n}}}\ \text{in expression}\ \displaystyle\sqrt{{{\left({x}+{1}\right)}^{2}}}.$$
Therefore,
$$\displaystyle\sqrt{{{\left({x}+{1}\right)}^{2}}}={\left({\left({x}+{1}\right)}^{2}\right)}^{{\frac{1}{{2}}}}$$
$$\displaystyle={\left({x}+{1}\right)}^{{{2}\cdot\frac{1}{{2}}}}$$
$$\displaystyle={\left({x}+{1}\right)}^{1}$$
$$\displaystyle={\left({x}+{1}\right)}$$
Therefore,
$$\displaystyle\frac{\sqrt{{12}}}{\sqrt{{{x}+{1}}}}=\frac{{{\sqrt[{{2}}]{{3}}}\sqrt{{{x}+{1}}}}}{{{x}+{1}}}$$
Apply the product property of radicals: $$\displaystyle{\sqrt[{{n}}]{{a}}}{\sqrt[{{n}}]{{b}}}={\sqrt[{{n}}]{{{a}{b}}}}$$ in the numerator,
$$\displaystyle\sqrt{{3}}\sqrt{{{x}+{1}}}=\sqrt{{{3}{\left({x}+{1}\right)}}}$$
$$\displaystyle=\sqrt{{{3}{x}+{3}}}$$
Hence,
$$\displaystyle\frac{\sqrt{{12}}}{\sqrt{{{x}+{1}}}}=\frac{{\sqrt[{{2}}]{{{3}{x}+{3}}}}}{{{x}+{1}}}$$
Therefore, the simplified form of the expression $$\displaystyle\frac{\sqrt{{12}}}{\sqrt{{{x}+{1}}}}\text{ is }\frac{{\sqrt[{{2}}]{{{3}{x}+{3}}}}}{{{x}+{1}}}.$$