Question

How do you simplify \[\sqrt {68} \]?

Answer 1

Hint: In the given question, we have been asked to find the simplified form of the square root of an even natural number. To solve this question, we just need to know how to solve the square root. If the number is a perfect square, then it will have no integer left in the square root. But if it is not a perfect square, then it has at least one integer in the square root.

Complete step-by-step answer:
The given number whose simplified form is to be found is the square root of \[68\], or we have to evaluate the value of \[\sqrt {68} \].
First, we find the prime factorization of \[68\] and club the pair(s) of equal integers together.
\[\begin{array}{l}{\rm{ }}2\left| \!{\underline {\,
  {68} \,}} \right. \\{\rm{ }}2\left| \!{\underline {\,
  {34} \,}} \right. \\17\left| \!{\underline {\,
  {17} \,}} \right. \\{\rm{ }}\left| \!{\underline {\,
  1 \,}} \right. \end{array}\]
Hence, \[68 = 2 \times 2 \times 17 = {2^2} \times 17\]
Hence, \[\sqrt {68} = \sqrt {{{\left( 2 \right)}^2} \times 17} = 2\sqrt {17} \]
Thus, the simplified form of \[\sqrt {68} \] is \[2\sqrt {17} \].

Note:When we are calculating for such questions, we find the prime factorization, club the pairs together, take them out as a single number and solve for it. This procedure requires no further action or steps to evaluate the answer. It is a point to remember that a perfect square always has an even number of factors.