The question asks to simplify a radical which is 4 - 64 and this equals radical 60 and simplified it equals 2 radical 15.
in Trigonometry Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

To radicalise a square root you break the number down to its factors and then decide whether you can pair any of the factors. For example, if the number is 64, this can be broken down to 2*2*2*2*2*2, that is 3 pairs of twos. When you take the square root, each pair reduces to one single number: so we have (2*2)(2*2)(2*2) reducing to 2*2*2. This is the number 8.

But it's not always that easy. Another example, 180=2*2*3*3*5=(2*2)(3*3)5. This time we can reduce the pairs to 2*3=6, but we have a 5 left. To radicalise the square root we take the reduced pairs, 6, and multiply by the square root of 5: 6sqrt(5). That is square root of 180 in radical form.

Now let's take 60=2*2*3*5=(2*2)3*5. Square root is 2sqrt(15). Example: 1350=2*3*3*3*5*5=2(3*3)3(5*5). We can reduce the pairs to 3*5 but we're left with 2*3=6; so the square root in radical form is 3*5sqrt(6)=15sqrt(6).

We can't radicalise negative numbers unless we involve the imaginary number i, defined as the square root of -1. But  if we wanted to radicalise square root of -60, we would radicalise 60 then follow it with i: 2sqrt(15)i or 2isqrt(15).

Not all numbers can be radicalised, but that just means we can't pair any of its factors.

Other roots, like cube roots, can be radicalised, too, but instead of pairs we group in threes, so cube root of 1350=2(3*3*3)5*5. Reduce the group of three 3s to just one: 3cuberoot(50).  But 64=(2*2*2)(2*2*2), so the cube root of 64 is 2*2=4 because each of three 2s reduces to 2. For fourth roots we group in fours, and so on.

Radicalisation happens in algebra, too. 25(x+1)(x-2)^2= (5*5)((x-2)*(x-2))(x+1), so square root is 5(x-2)sqrt(x+1).

by Top Rated User (1.1m points)

Related questions

2 answers
asked Sep 30, 2014 in Algebra 2 Answers by lovelymath Level 2 User (1.9k points) | 698 views
1 answer
asked Sep 30, 2014 in Algebra 2 Answers by lovelymath Level 2 User (1.9k points) | 620 views
1 answer
asked Sep 30, 2014 in Algebra 2 Answers by lovelymath Level 2 User (1.9k points) | 658 views
3 answers
asked Sep 30, 2014 in Algebra 1 Answers by lovelymath Level 2 User (1.9k points) | 865 views
1 answer
asked Apr 1, 2013 in Algebra 2 Answers by anonymous | 607 views
1 answer
asked Sep 1, 2014 in Algebra 2 Answers by anonymous | 609 views
1 answer
asked May 17, 2014 in Other Math Topics by HARSH RATHI | 574 views
1 answer
asked May 17, 2014 in Other Math Topics by HARSH RATHI | 539 views
1 answer
1 answer
asked Aug 11, 2013 in Algebra 2 Answers by Jennifer A. Cascaño Level 12 User (101k points) | 516 views
1 answer
asked Aug 10, 2013 in Algebra 2 Answers by Jennifer A. Cascaño Level 12 User (101k points) | 584 views
1 answer
asked Aug 10, 2013 in Algebra 2 Answers by Jennifer A. Cascaño Level 12 User (101k points) | 712 views
1 answer
asked Jul 20, 2013 in Algebra 2 Answers by Jennifer A. Cascaño Level 12 User (101k points) | 662 views
1 answer
asked Jul 20, 2013 in Algebra 2 Answers by Jennifer A. Cascaño Level 12 User (101k points) | 646 views
1 answer
asked Jul 20, 2013 in Algebra 2 Answers by Jennifer A. Cascaño Level 12 User (101k points) | 589 views
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,542 questions
99,804 answers
2,417 comments
523,353 users