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avatar+581 

If -3 < a < 4, and -2 < b < 2, then what is the range of (a+b)^2 ?

 

A) 0 ≤ (a+b)^2 < 36

B) 25 < (a+b)^2 < 36

C) -25 < (a+b)^2 < 36

D) -36 < (a+b)^2 < 25

E) o < (a+b)^2 < 36

 

So apparently the correct answer is A. I need some clarification.

Please explain the process. Thanks

 Sep 8, 2019
 #1
avatar+104723 
+2

Note that   (a + b)^2   either produces a positive result   or  0

 

So.....the least that (a + b)^2  could be is  0

 

And the greatest that it can be  is if  a ≈ 4  and b  ≈ 2

 

So    (a + b)^2  =  ( ≈4  + ≈2) ^2   =  (≈ 6)^2  =  ≈ 36

 

Therefore

 

0 ≤ (a+b)^2 < 36

 

 

cool cool cool

 Sep 8, 2019
 #2
avatar+581 
+1

Ah thanks CPhill. You've been saving me for the past few questions. I really appreciate it. 

 

Just a quick follow up. Can you explain why it can't be option B?

Gh0sty15  Sep 8, 2019
edited by Gh0sty15  Sep 8, 2019
 #3
avatar+104723 
+1

It can't be "B"  because we need the entire  range possible  of  (a+b)^2

 

"B"  says  that   (a + b)^2   must be  >  25    [ but   < 36 ]

 

But this isn't  true......note  if   a = 0   and  b  = 0....then (a + b)^2  = (0 + 0)^2  =  0

 

So....we can make   (a +b)^2   less than 25

 

 

cool cool cool

CPhill  Sep 8, 2019

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