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For how many positive integers $x$ is $x^2 + 6x + 9$ between 20 and 40?

 Jun 15, 2019
 #1
avatar+8406 
+3

This probably isn't the most sophisticated way, but we can easily solve this just by plugging in different numbers.

 

Let   f(x)  =  x2 + 6x + 9

 

f(1)  =  12 + 6(1) + 9  =  1 + 6 + 9  =  16

 

f(2)  =  22 + 6(2) + 9  =  4 + 12 + 9  =  25

 

f(3)  =  32 + 6(3) + 9  =  9 + 18 + 9  =  36

 

f(4)  =  42 + 6(4) + 9  =  16 + 24 + 9  =  49

 

So we can see that there are only  2  positive integers  x  that satisfy  20 < f(x) < 40

 

They are:     x = 2     and     x = 3

 Jun 15, 2019
 #2
avatar+101856 
+2

x^2 + 6x + 9  = 20                              x^2 + 6x + 9  = 40                              

(x + 3)^2  = 20                                   (x + 3)^2  = 40

x + 3  = √20                                        x + 3 = √40

x = √20 - 3                                          x  = √40 - 3

x = ceiling [ √20 - 3 ]  = 2                    x = floor [ √40 - 3 ]  = 3

 

As hectictar found....2 and 3 fit the bill....!!!!!

 

 

cool cool cool

 Jun 15, 2019

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