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slava [35]
3 years ago
8

Explain how to get that answer!!

Mathematics
1 answer:
ra1l [238]3 years ago
6 0
We need to simplify \frac{ \sqrt{14x^3} }{ \sqrt{18x} }

First lets factor \sqrt{14x^3}

\sqrt{14x^3} = \sqrt{14}  \sqrt{x^3}
\sqrt{14} =  \sqrt{2} \sqrt{7} by applying the radical rule \sqrt[n]{ab} =  \sqrt[n]{a} \sqrt[n]{b}
\sqrt{x^3} = x^{3/2} By applying the radical rule \sqrt[n]{x^m} = x^{m/n}

So
\sqrt{14x^3} = \sqrt{14}  \sqrt{x^3} = \sqrt{2} \sqrt{7}x^{3/2}

Now let's factor \sqrt{18x}
By applying the radical rule \sqrt[n]{ab} =  \sqrt[n]{a}  \sqrt[n]{b},
\sqrt{18x} =  \sqrt{18} \sqrt{x}
\sqrt{18} =  \sqrt{2} * 3

So \sqrt{18x} = \sqrt{2}*3 \sqrt{x}

So  \frac{ \sqrt{14x^3} }{ \sqrt{18x} } = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}  }

We know that \sqrt[n]{x} = x^{1/n} so \sqrt{x} = x^{1/2}

We now have \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}

We know that \frac{x^a}{x^b} = x^{a-b}
So \frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x

We now got \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}


We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} =   \frac{ \sqrt{7}x }{3}


All in All, we get \frac{ \sqrt{14x^3} }{ \sqrt{18x} } =  \frac{ \sqrt{7}x }{3}

Hope this helps! :D


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