Home > Algebra > Square Roots and Irrational Numbers

How do you simplify #sqrt(1215)#?

Answer 1

Eli Assouline

#9sqrt(15)#

First decompose 1215 in prime factors:

The last digit is 5 so it is multiple o f5:

#1215/5=243#

The sum of the digits of 243 its is 9, therefore it is multiple 3.

#243/3=81#.

We know by the grammar school that #81=9xx9=3^4#.

So #1215=3^5*5#

#Sqrt(1215)=sqrt(3^5*5)=sqrt(3^4*3*5)=3^2sqrt(3*5)=9sqrt(15)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Eli Assouline

To simplify √1215, you can factor 1215 into its prime factors:

1215 = 3 * 3 * 3 * 3 * 3 * 3 * 5

Now, since there are three pairs of identical factors of 3, you can take them out of the square root:

√1215 = 3 * 3 * 3 * √5

Simplify the expression under the square root:

√1215 = 27√5

Therefore, the simplified form of √1215 is 27√5.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.