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

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