Question

What is a square root of 5 times the square root of 35?

Answer 1

Hint: We try to form the indices formula for the value 2. This is a multiplication of the square root of the multiplication of 5 and the square root of 35. We find the prime factorisation of 35. Then we take indices form of the multiplication. We complete the multiplication in the place of those indices.

Complete step by step solution:
We need to find the value of the multiplication of the square root of 5 times the square root of 35. This is a square root form.
The given value is the form of indices. We are trying to find the root value of 35.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 2 we get \[{{a}^{\dfrac{1}{2}}}=\sqrt[2]{a}=\sqrt{a}\].
We also know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$
We need to find the prime factorisation of the given number 35.
$\begin{align}
  & 5\left| \!{\underline {\,
  35 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore, \[35=5\times 7\].
Therefore, \[\sqrt[2]{35}={{35}^{\dfrac{1}{2}}}={{5}^{\dfrac{1}{2}}}{{7}^{\dfrac{1}{2}}}\]. We now multiply 5 with \[{{5}^{\dfrac{1}{2}}}{{7}^{\dfrac{1}{2}}}\]
5 can be represented as \[5={{5}^{1}}\].
We have the identity formula of ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$.
Therefore, \[5\times {{5}^{\dfrac{1}{2}}}{{7}^{\dfrac{1}{2}}}={{5}^{1+\dfrac{1}{2}}}{{7}^{\dfrac{1}{2}}}={{5}^{\dfrac{3}{2}}}{{7}^{\dfrac{1}{2}}}\].
Now we need to find the square root of \[5\times {{5}^{\dfrac{1}{2}}}{{7}^{\dfrac{1}{2}}}={{5}^{1+\dfrac{1}{2}}}{{7}^{\dfrac{1}{2}}}={{5}^{\dfrac{3}{2}}}{{7}^{\dfrac{1}{2}}}\].
We can write the mathematical form as \[\sqrt{{{5}^{\dfrac{3}{2}}}{{7}^{\dfrac{1}{2}}}}={{\left( {{5}^{\dfrac{3}{2}}}{{7}^{\dfrac{1}{2}}} \right)}^{\dfrac{1}{2}}}\].
We again apply the identity formula of ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$.
Therefore, \[{{\left( {{5}^{\dfrac{3}{2}}}{{7}^{\dfrac{1}{2}}} \right)}^{\dfrac{1}{2}}}={{5}^{\dfrac{3}{2}\times \dfrac{1}{2}}}{{7}^{\dfrac{1}{2}\times \dfrac{1}{2}}}={{5}^{\dfrac{3}{4}}}{{7}^{\dfrac{1}{4}}}\].

The final solution is \[{{5}^{\dfrac{3}{4}}}{{7}^{\dfrac{1}{4}}}\].

Note: If we are finding the square and cube root of any numbers, we don’t always need to find the all-possible roots. The problem gets more complicated in that case. The simplification of the indices is the main criteria we need to find the simplified form of the given fraction.