How do you simplify #sqrt24*sqrt(80/192)#?

2 Answers
smendyka
Jul 17, 2017

See a solution process below:

Explanation:

First, rewrite this expression as:

#sqrt(4 * 6)sqrt((16 * 5)/(64 * 3))#

We can then use these rules for radicals to begin simplifying the expression:

#sqrt(color(red)(a)/color(blue)(b)) = sqrt(color(red)(a))/sqrt(color(blue)(b))# and #sqrt(color(red)(a) * color(blue)(b)) = sqrt(color(red)(a)) * sqrt(color(blue)(b))#

#sqrt(4 * 6)sqrt((16 * 5)/(64 * 3)) =>#

#sqrt(4 * 6)sqrt((16 * 5))/sqrt((64 * 3)) =>#

#sqrt(4)sqrt(6)(sqrt(16)sqrt(5))/(sqrt(64)sqrt(3)) =>#

#2sqrt(6)(4sqrt(5))/(8sqrt(3)) =>#

#(2 * 4/8)sqrt(6)sqrt(5)/sqrt(3) =>#

#(8/8)sqrt(6)sqrt(5)/sqrt(3) =>#

#1sqrt(6)sqrt(5)/sqrt(3) =>#

#sqrt(6)sqrt(5)/sqrt(3)#

We can now use the reverse of the rules above to complete the simplification process:

#sqrt(6)sqrt(5)/sqrt(3) =>#

#(sqrt(6)sqrt(5))/sqrt(3) =>#

#sqrt(6 * 5)/sqrt(3) =>#

#sqrt((6 * 5)/3) =>#

#sqrt(30/3) =>#

#sqrt(10)#

EZ as pi
Jul 17, 2017

#sqrt10#

Explanation:

Square roots which are multiplied or divided can be combined into one square root:

#sqrt24 xx sqrt(80/192) = sqrt((24xx80)/192)#

Write each number as the product of its factors, using square numbers where possible.

#sqrt((24xx80)/192) =sqrt((6xx4xx16xx5)/(3xx4xx16)#

Simplify:
#sqrt((cancel6^2xxcancel4xxcancel16xx5)/(cancel3xxcancel4xxcancel16)#

#= sqrt10#

If there were no square numbers, you would simply have used the prime factors which would give the same results

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
2949 views around the world
You can reuse this answer
Creative Commons License