Question

How do you simplify $\sqrt{320}$?

Answer 1

Hint: We try to form the indices formula for the value 2. This is a square root of 320. We find the prime factorisation of 320. Then we take one digit out of the two same number of primes. There will be some odd number of primes remaining in the root. We keep them as it is.

Complete step by step answer:
We need to find the value of the algebraic form of $\sqrt{320}$. This is a square root form.
The given value is the form of indices. We are trying to find the root value of 320.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 2 we get \[{{a}^{\dfrac{1}{2}}}=\sqrt{a}\].
We need to find the prime factorisation of the given number 320.
$\begin{align}
  & 2\left| \!{\underline {\,
  320 \,}} \right. \\
 & 2\left| \!{\underline {\,
  160 \,}} \right. \\
 & 2\left| \!{\underline {\,
  80 \,}} \right. \\
 & 2\left| \!{\underline {\,
  40 \,}} \right. \\
 & 2\left| \!{\underline {\,
  20 \,}} \right. \\
 & 2\left| \!{\underline {\,
  10 \,}} \right. \\
 & 1\left| \!{\underline {\,
  5 \,}} \right. \\
\end{align}$
Therefore, $320=2\times 2\times 2\times 2\times 2\times 2\times 5$.
For finding the square root, we need to take one digit out of the two same number of primes.
This means in the root value of $320=2\times 2\times 2\times 2\times 2\times 2\times 5$, we will take out three 2s but one five is an odd number of times. In that case we keep the number inside the root.
So, $\sqrt{320}=\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 5}=2\times 2\times 2\times \sqrt{5}=8\sqrt{5}$. Basically 320 is the square of $8\sqrt{5}$
We can also use the theorem of indices \[{{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}\]. We know that $320={{\left( 8\sqrt{5} \right)}^{2}}$.
We need to find $\sqrt{320}$ which gives $\sqrt{320}={{\left( 320 \right)}^{\dfrac{1}{2}}}={{\left[ {{\left( 8\sqrt{5} \right)}^{2}} \right]}^{\dfrac{1}{2}}}={{\left( 8\sqrt{5} \right)}^{2\times \dfrac{1}{2}}}=8\sqrt{5}$.
Therefore, the value of $\sqrt{320}$ is $8\sqrt{5}$.

Note: We can also use the variable form where we can take $x=\sqrt{320}$. But we need to remember that we can’t use the square on both sides of the equation $x=\sqrt{320}$ as in that case we are taking an extra value of negative as in $-8\sqrt{5}$ as a root value. Then this linear equation becomes a quadratic equation.