Question

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

Rational exponents and radicals
ANSWERED
asked 2020-12-16
To calculate: The simplified form of the expression \(\displaystyle\frac{\sqrt{{12}}}{\sqrt{{{x}+{1}}}}.\)

Answers (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 \(\displaystyle\frac{\sqrt{{12}}}{\sqrt{{{x}+{1}}}}\text{ is }\frac{{\sqrt[{{2}}]{{{3}{x}+{3}}}}}{{{x}+{1}}}.\)

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