Question

What is 9 square root of 40 simplified?

Answer 1

Hint: We try to form the indices formula for the value 2. This is a cube root of 40. We find the prime factorisation of 40. Then we take one digit out of the two same number of primes. There will be two odd numbers of primes remaining in the root which can’t be taken out. We keep the cube root in its simplest form and multiply with 9.

Complete step by step answer:
We need to find the value of the algebraic form of $\sqrt[2]{40}$. This is a square root form.
The given value is the form of indices. We are trying to find the root value of 40.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 2 we get \[{{a}^{\dfrac{1}{2}}}=\sqrt[2]{a}\].
We need to find the prime factorisation of the given number 40.
$\begin{align}
  & 2\left| \!{\underline {\,
  40 \,}} \right. \\
 & 2\left| \!{\underline {\,
  20 \,}} \right. \\
 & 2\left| \!{\underline {\,
  10 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore, \[40=2\times 2\times 2\times 5\].
For finding the square root, we need to take one digit out of the two same number of primes.
This means in the square root value of \[40=2\times 2\times 2\times 5\], we will take out one 2 from the multiplication. One 2 and one 5 will remain in the root.
So, \[\sqrt[2]{40}=\sqrt[2]{2\times 2\times 2\times 5}=2\sqrt{10}\]. Basically 40 is the square of \[2\sqrt{10}\].
Now we multiply 9 with \[2\sqrt{10}\] and get \[9\times 2\sqrt{10}=18\sqrt{10}\].

Therefore, 9 square root of 40 simplified is \[18\sqrt{10}\].

Note: We can also use the variable form where we can take $x=\sqrt[2]{40}$. But we need to remember that we can’t use the cube on both sides of the equation $x=\sqrt[2]{40}$ as in that case we are taking two extra values as a root value. Then this linear equation becomes a cubic equation.