Question

What is the square root of 6 times the square root of 2?

Answer 1

Hint: We try to form the indices formula for the value 2. This is a multiplication of square root of 6 and square root of 2. We find the prime factorisation of 6. Then we take one digit out of the two same number of primes from the multiplication. There will be an odd number of primes remaining in the root which can’t be taken out. We keep the square root in its simplest form.

Complete step-by-step solution:
We need to find the value of the multiplication of the square root of 2 times the square root of 6. This is a square root form.
The given value is the form of indices. We are trying to find the root value of 6.
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 need to find the prime factorisation of the given number 6.
$\begin{align}
  & 2\left| \!{\underline {\,
  6 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore, \[6=2\times 3\].
For finding the square root, we need to take one digit out of the two same number of primes.
The multiplied terms are \[\sqrt{2}\] and \[\sqrt{6}\].
So, the multiplication is \[\sqrt{2}\times \sqrt{6}=\sqrt{2\times 2\times 3}\].
This means in the square root value we will take out two fives from the multiplication.
So, \[\sqrt{2}\times \sqrt{6}=\sqrt{2\times 2\times 3}=2\sqrt{3}\].

Note: We can also use the variable form where we can take $x=\sqrt[2]{12}$. But we need to remember that we can’t use the cube on both sides of the equation $x=\sqrt[2]{12}$ as in that case we are taking two extra values as a root value. Then this linear equation becomes a cubic equation.