How do you simplify #2 sqrt20 + 8 sqrt45 - sqrt80#?

1 Answer
Apr 15, 2018

Answer:

The answer is #24sqrt(5)#.

Explanation:

Note: when the variables a, b, and c are used, I am referring to a general rule that will work for every real value of a, b, or c.

You can use the rule #sqrt(a*b)=sqrt(a)*sqrt(b)# to your advantage:

#2sqrt(20)# equals #2sqrt(4*5)#, or #2sqrt(4)*sqrt(5)#.

Since #sqrt(4)=2#, you can substitute #2# in to get #2*2*sqrt(5)#, or #4sqrt(5)#.

Use the same rule for #8sqrt(45)# and #sqrt(80)#:

#8sqrt(45) -> 8sqrt(9*5) -> 8sqrt(9)*sqrt(5) -> 8*3*sqrt(5) -> 24sqrt(5)#.

#sqrt(80) -> sqrt(16*5) -> sqrt(16)*sqrt(5) -> 4sqrt(5)#.

Substitute these into the original equation and you get:

#4sqrt(5) + 24sqrt(5) - 4sqrt(5)#.

Since #asqrt(c)+bsqrt(c)=(a+b)sqrt(c)#, and likewise #asqrt(c)-bsqrt(c)=(a-b)sqrt(c)#, you can simplify the equation:

#4sqrt(5) + 24sqrt(5) - 4sqrt(5) -> 28sqrt(5)-4sqrt(5) -> 24sqrt(5)# , the final answer.

Hope this helps!