How do you evaluate 4√112 +5√56 - 9√126?

4 Answers
Jul 9, 2018

Answer:

#16sqrt(7)-17sqrt(14)#

Explanation:

Note that

#112=7*16#

#56=4*14#

#126=9*14#

and

#sqrt(ab)=sqrt(a)*sqrt(b)# if #a,b>=0#
so we get

#16sqrt(17)+10sqrt(14)-27sqrt(14)#

combining like Terms

#16sqrt(7)-17sqrt(14)#

Jul 9, 2018

Answer:

#16sqrt7-17sqrt14#

Explanation:

Write each radicand (the number under the root) as a product of its factors. Try to use perfect squares wherever possible.

#4sqrt112+5sqrt56-9sqrt126#

#= 4sqrt(16xx7) +5sqrt(4xx2xx7) -9sqrt(9xx14)#

Find any roots possible:

#=4xx4xxsqrt7 +5 xx2 sqrt(2xx7)-9xx3sqrt(2xx7)#

#=16sqrt7 +10sqrt14-27sqrt14#

#=16sqrt7-17sqrt14" "larr# this can be the answer

Or, we can try to improve it a bit

#=16sqrt7-17sqrt7sqrt2#

This can be factored to give:

#sqrt7(16-17sqrt2)#

However, this answer is no better than the first.

Jul 9, 2018

Answer:

#16sqrt7-17sqrt14#

Explanation:

We have the following:

#4sqrt112+5sqrt56-9sqrt126#

In each of our terms, we can rewrite the radicals in such a way where we can factor out a perfect square:

#color(blue)(4sqrt(16*7))+color(purple)(5sqrt(4*14))-color(steelblue)(9sqrt(9*14))#

This business simplifies to

#color(blue)(16sqrt7)+color(purple)(10sqrt14)-color(steelblue)(27sqrt14)#

Since the latter two terms have a #sqrt14# in common, we can simplify those to get

#color(blue)(16sqrt7)-color(red)(17sqrt14)#

Since the radicals have no perfect square factors, we are done!

Hope this helps!

Jul 9, 2018

Answer:

#16sqrt7-17sqrt14#

Explanation:

#4sqrt112+5 sqrt 56-9 sqrt 126#

#:.=4 sqrt 7*sqrt 16+5 sqrt 7* sqrt 8-9 sqrt(2*3*3*7)#

#:.=4*4 sqrt 7+5 *sqrt 7*sqrt 8-27*sqrt 2*sqrt 7#

#:.=sqrt 7(16+5 sqrt(2*2*2)-27 sqrt 2)#

#:.=sqrt 7(16+10 sqrt 2-27 sqrt 2)#

#:.=sqrt 7(16+sqrt 2(10-27))#

#:.=sqrt 7(16+sqrt 2(-17))#

#:.=sqrt 7(16-17 sqrt 2)#

#:.=16 *sqrt 7-17 sqrt 2*sqrt 7#

#:.=16sqrt 7-17 sqrt 14#