Adding and subtracting radicals 2x\sqrt{3x^2y}\cdot3\sqrt{15xy^3}

cistG 2021-09-23 Answered
Adding and subtracting radicals \(\displaystyle{2}{x}\sqrt{{{3}{x}^{{2}}{y}}}\cdot{3}\sqrt{{{15}{x}{y}^{{3}}}}\)

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Expert Answer

Leonard Stokes
Answered 2021-09-24 Author has 7005 answers
Given
\(\displaystyle{2}{x}\sqrt{{{3}{x}^{{2}}{y}}}\cdot{3}\sqrt{{{15}{x}{y}^{{3}}}}\)
We have to solve for the simple form
\(\displaystyle{\left({2}{x}\right)}{\left({3}\right)}\cdot\sqrt{{{3}{x}^{{2}}{y}}}\cdot\sqrt{{{15}{x}{y}^{{3}}}}\)
\(\displaystyle={6}{x}\cdot\sqrt{{{\left({3}{x}^{{2}}{y}\right)}\cdot{\left({15}{x}{y}^{{3}}\right)}}}\)
\(\displaystyle={6}{x}\cdot\sqrt{{{\left({3}\right)}{\left({15}\right)}{\left({x}^{{2}}\right)}{\left({x}\right)}{\left({y}\right)}{\left({y}^{{3}}\right)}}}\)
\(\displaystyle={6}{x}\cdot\sqrt{{{45}\cdot{x}^{{{2}+{1}}}\cdot{y}^{{{1}+{3}}}}}\)
\(\displaystyle={6}{x}\cdot\sqrt{{{45}\cdot{x}^{{3}}\cdot{y}^{{4}}}}\)
\(\displaystyle={6}{x}\sqrt{{{\left({9}\times{5}\right)}\cdot{x}^{{3}}\cdot{\left({y}^{{2}}\right)}^{{2}}}}\)
\(\displaystyle={6}{x}\sqrt{{{\left({9}\times{5}\right)}\cdot{x}^{{3}}\cdot{\left({y}^{{2}}\right)}^{{2}}}}\)
\(\displaystyle={6}{x}\sqrt{{{\left({3}^{{2}}\times{5}\right)}\cdot{x}^{{3}}\cdot{\left({y}^{{2}}\right)}^{{2}}}}\)
\(\displaystyle={6}{x}\sqrt{{{\left({3}{y}^{{2}}\right)}^{{2}}\cdot{5}{x}^{{3}}}}\)
\(\displaystyle={6}{x}\cdot{\left({3}{y}^{{2}}\right)}\cdot\sqrt{{{5}{x}^{{3}}}}\)
\(\displaystyle={18}{x}{y}^{{2}}\cdot\sqrt{{{5}{x}^{{3}}}}\)
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