How do you order the following from least to greatest #4 4/5, sqrt19, -sqrt5, -21, sqrt5#?

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Answer 1 · 2017-01-19T23:28:48

#-21<-sqrt(5)< sqrt(5)<4 4/5 < sqrt(19)#

Notice that #-21# and #-sqrt(5)# are negative

or

#-21<0# and #-sqrt(5)<0#

One way to think of this when it comes to the order of these two negatives is that

#-21<-20# therefore #-sqrt(21)<-sqrt(20)=-(sqrt(4(5)))=-(sqrt(4)(sqrt(5)))=-2sqrt(5)#

So

#-21<-2sqrt(5)< -sqrt(5)#

Then

#-21< -sqrt(5)#

That takes care of the negative numbers, but we still have the three positive numbers.

First notice

#4 4/5= 16/5=3.5#

and also

#4<5<9#

#<=>#

#sqrt(4)< sqrt(5)< sqrt(9)# Take the square root of all sides

#<=>#

#2< sqrt(5) < 3#

Therefore #sqrt(5)< 4 4/5#

#4 4/5=3.5 => (4 4/5)^2=(3.5)^2#

#=(3+0.5)^2=9+2(3)(0.5)+0.25=9+3+0.25=12.25#

So we know that

#4 4/5=sqrt(12.25)#

and since

#12.25<19#

then

#sqrt(12.25) < sqrt(19)#

So we get

#ul(-21 < -sqrt(5) < sqrt(5) <4 4/5 < sqrt(19))#