How do you simplify #sqrt(20d) + 4sqrt(12d) - 3 sqrt(45d)#?

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Answer 1 · 2017-08-15T18:27:31

#=8sqrt(3d) -7sqrt(5d)#

Write each radicand as the product of its factors and try to find squares wherever possible:

#sqrt(20d)+4sqrt(12d) -3sqrt(45d)#

#=sqrt(4xx5d)+4sqrt(4xx3d) -3sqrt(9xx5d)" "larr# find the roots

#=2sqrt(5d)+4xx2sqrt(3d) -3xx3sqrt(5d)#

#=2sqrt(5d)+8sqrt(3d) -9sqrt(5d)" "larr# collect like terms

#=8sqrt(3d) -7sqrt(5d)#

Answer 2 · 2017-08-15T18:29:02

After simplifying, you are left with #8sqrt(3d)-7sqrt(5d)#

First, let's simplify the numbers under the square root.

#sqrt(20)#

This can be rewritten as #sqrt(4*5)#

4 has a square root, so we can pull its square root out of the radical, giving us #2sqrt(5)#

Now we do the same thing to #sqrt(12)# and #sqrt(45)#

#sqrt(12)=sqrt(4*3)=2sqrt(3)#

#sqrt(45)=sqrt(9*5)=3sqrt(5)#

We can't do much with the #sqrt(d)#, so at this point we have:

#2sqrt(5d)+4*2sqrt(3d)-3*3sqrt(5d)#

After multiplication, we have:

#2sqrt(5d)+8sqrt(3d)-9sqrt(5d)#

We have two #sqrt(5d)# terms, so we can combine them, giving us the final answer of:

#8sqrt(3d)-7sqrt(5d)#