How do you simplify #sqrt(75/4 )#?

2 Answers
Jun 11, 2018

Answer:

#\frac{5sqrt(3)}{2}#

Explanation:

When you need to simplify numerical root, you should look for these two key facts:

  • #sqrt(a^2) = a#
  • #sqrt(ab)=sqrt(a)sqrt(b)#

So, anytime you have a number inside a root, you should try to write it as a product of other numbers, of which at least one is a perfect square. Let's analyze your case.

First of all, using the second property, we can write

#sqrt(75/4) = sqrt(75)/sqrt(4)#

In fact, every fraction can be read as multiplication using

#\frac{a}{b} = a * \frac{1}{b}#

Now let's deal with the two roots separately: I'd start with the denominator, since we already have a perfect square under a square root, so they simplify (see first rule above): we have

#sqrt(75) / sqrt(4) = sqrt(75)/2#

As for #sqrt(75)#, we can see that #75 = 25*3#, and #25=5^2# is a perfect square. So, by the second rule above, we have

#sqrt(75) = sqrt(25*3) =sqrt(25)*sqrt(3)=5sqrt(3)#

Which leads to the final answer

#sqrt(75/4) = \frac{5sqrt(3)}{2}#

But how do we find the most appropriate way to rewrite our number, in this case #75=25*3#? You can use the prime factorization:

#75 = 3*5^2#

and select only the primes with even exponent. In this case, #5^2#.

Jun 11, 2018

Answer:

# +-(5sqrt(3))/2#

Explanation:

If you are stuck it is always worth have in a 'play' with numbers and see what comes up.

Consider the 75. This is exactly divisible by 5 so #75-:5=15#

Thus we have: #sqrt((5xx15)/4)#

But 15 is the same as #3xx5# giving:

#sqrt((5xx5xx3)/4)#

#sqrt(5^2xx3)/sqrt(4) = +-(5sqrt(3))/2#