Math Question

It says solve the problem using these words: product, rule, principle, root, quotient rule, reciprical, nth root. 

1. √b^5

2. √2s^5* √10s^4    Thanks!

I guess everything behind your root is supposed to be the radical.

1.

\(\sqrt[2]{b^5}=(b^5)^{\frac{1}{2}}\)   ⇒   \(\sqrt[n]{x} =x^\frac{1}{n}\)                                                                                                 The exponent of the power is the reciprocal of the root exponent.

\(\sqrt{b^5}=\sqrt{b\times b\times b\times b\times b}\)  \(=\sqrt{b\times b^4}=b^2\sqrt{b}\)  .

The exponent of the result is the quotient of the exponent of the radiculor and the root-exponent.   ⇒   \(\sqrt[2]{b^4} =b^{\frac{4}{2}}= b^2\)

2.

\(\sqrt{2s^5}\times \sqrt{10s^4}=\sqrt{(2\times s^\times s^4)\times (2\times 5\times s^4) }\)

Roots with the same root exponents are grouped under a single root.

\( =\sqrt{s \times 5\times2^2\times s^4\times s^4}\)

\(= 2s^4\sqrt{5s}\)

   !

It says solve the problem using these words: product, rule, principle, root, quotient rule, reciprical, nth root. 

1. √b^5

\(\sqrt{b}^5=(b^{1/2})^5=b^{5/2}=b^{2.5}\\ or\\ \sqrt{b^5}=(b^5)^{1/2}=b^{5/2}=b^{2.5}\)

2. √2s^5* √10s^4

\(\sqrt2*s^5*\sqrt{10}*s^4\\ =\sqrt{20}*s^9\\ =\sqrt{4*5}*s^9\\ =2\sqrt{5}\;s^9\\ \)