Author 
Message 
TAGS:

Hide Tags

Retired Moderator
Joined: 25 Feb 2013
Posts: 1167
Location: India
GPA: 3.82

If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
Show Tags
Updated on: 18 Sep 2017, 09:25
Question Stats:
53% (02:14) correct 47% (02:26) wrong based on 226 sessions
HideShow timer Statistics
If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true A.) \(g(x) < b\) B.) \(g(x) < 0\) C.) \(0 < g(x) < 4b\) D.) \(g(x) ≥ 4b\) E.) \(g(x) ≥ 5b^2\)
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by niks18 on 17 Sep 2017, 08:04.
Last edited by niks18 on 18 Sep 2017, 09:25, edited 1 time in total.



Current Student
Joined: 22 Apr 2017
Posts: 106
Location: India
GMAT 1: 620 Q46 V30 GMAT 2: 620 Q47 V29 GMAT 3: 630 Q49 V26 GMAT 4: 690 Q48 V35
GPA: 3.7

Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
Show Tags
18 Sep 2017, 09:32
niks18 wrote: If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true
A.) \(g(x) < b\)
B.) \(g(x) < 0\)
C.) \(0 < g(x) < 4b\)
D.) \(g(x) ≥ 4b\)
E.) \(g(x) ≥ 5b^2\) g(x)= x^24bx+9b^2> (x2b)^2+5b^2. Since b is a positive integer, value of g(x) will always be greater than 5b^2 as (x2b)^2 will alwz be positive. Hence for min value of b i.e 1, g(x)=5b^2. Hence E. g(x)>=5b^2



Retired Moderator
Joined: 25 Feb 2013
Posts: 1167
Location: India
GPA: 3.82

Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
Show Tags
18 Sep 2017, 11:11
niks18 wrote: If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true
A.) \(g(x) < b\)
B.) \(g(x) < 0\)
C.) \(0 < g(x) < 4b\)
D.) \(g(x) ≥ 4b\)
E.) \(g(x) ≥ 5b^2\) OE\(g(x) = x^2 – 4bx + 4b^2+5b^2\) or, \(g(x) = (x – 2b)^2+5b^2\), Now, \((x2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get \((x2b)^2+5b^2≥5b^2\) Or, \(g(x)≥5b^2\) Option E



Manager
Joined: 06 Aug 2017
Posts: 78
GMAT 1: 570 Q50 V18 GMAT 2: 610 Q49 V24 GMAT 3: 640 Q48 V29

Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
Show Tags
23 Sep 2017, 03:27
niks18 wrote: If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true
A.) \(g(x) < b\)
B.) \(g(x) < 0\)
C.) \(0 < g(x) < 4b\)
D.) \(g(x) ≥ 4b\)
E.) \(g(x) ≥ 5b^2\) Answer is E as follows. \(g(x) = x^2 – 4bx + 9b^2\) Since the above is a quadratic equation, its maxima/minima will be at ()4d/2 = 2b. Also since b is a +ve integer 2b will be +ve. Hence as per the rule the quadratic equation will have minimum value at x=2b. Putting the value of 2b back into the equation will yield g(2b) = \((2b)^2\)  4b*(2b) + 9\(b^2\) = 5\(b^2\)
_________________
 Kudos are the only way to tell whether my post is useful.
GMAT PREP 1: Q50 V34  700
Veritas Test 1: Q43 V34  630 Veritas Test 2: Q46 V30  620 Veritas Test 3: Q45 V29  610 Veritas Test 4: Q49 V30  650
GMAT PREP 2: Q50 V34  700
Veritas Test 5: Q47 V33  650 Veritas Test 5: Q46 V33  650



Senior Manager
Joined: 08 Jun 2015
Posts: 418
Location: India
GMAT 1: 640 Q48 V29 GMAT 2: 700 Q48 V38
GPA: 3.33

Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
Show Tags
28 Apr 2018, 22:24
+1 for option E. The given expression can be expressed as (x2b)^2+5b^2. The first part is always positive. The expression then is always greater than 5(b^2). Hence option E it is.
_________________
" The few , the fearless "



Intern
Joined: 15 Jan 2016
Posts: 36

Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
Show Tags
07 May 2018, 20:02
niks18 wrote: niks18 wrote: If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true
A.) \(g(x) < b\)
B.) \(g(x) < 0\)
C.) \(0 < g(x) < 4b\)
D.) \(g(x) ≥ 4b\)
E.) \(g(x) ≥ 5b^2\) OE\(g(x) = x^2 – 4bx + 4b^2+5b^2\) or, \(g(x) = (x – 2b)^2+5b^2\), Now, \((x2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get \((x2b)^2+5b^2≥5b^2\) Or, \(g(x)≥5b^2\) Option ECan you please further explain the need to factor the equation? I selected E because 9b^2 is the original equation. Any help would be super appreciated. Thanks Posted from my mobile device



Retired Moderator
Joined: 25 Feb 2013
Posts: 1167
Location: India
GPA: 3.82

Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
Show Tags
07 May 2018, 22:19
syedazeem3 wrote: niks18 wrote: niks18 wrote: If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true
A.) \(g(x) < b\)
B.) \(g(x) < 0\)
C.) \(0 < g(x) < 4b\)
D.) \(g(x) ≥ 4b\)
E.) \(g(x) ≥ 5b^2\) OE\(g(x) = x^2 – 4bx + 4b^2+5b^2\) or, \(g(x) = (x – 2b)^2+5b^2\), Now, \((x2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get \((x2b)^2+5b^2≥5b^2\) Or, \(g(x)≥5b^2\) Option ECan you please further explain the need to factor the equation? I selected E because 9b^2 is the original equation. Any help would be super appreciated. Thanks Posted from my mobile devicehi syedazeem3will be glad to clarify your doubts. I would like to understand how you deduced option E simply by looking at \(9b^2\) in the original equation. If instead of \(5b^2\), option E had mentioned \(4b^2\), then would you still have chosen option E as correct?



Intern
Joined: 15 Jan 2016
Posts: 36

Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
Show Tags
08 May 2018, 05:05
niks18 wrote: syedazeem3 wrote: niks18 wrote: [quote="niks18"]If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true
A.) \(g(x) < b\)
B.) \(g(x) < 0\)
C.) \(0 < g(x) < 4b\)
D.) \(g(x) ≥ 4b\)
E.) \(g(x) ≥ 5b^2\) OE\(g(x) = x^2 – 4bx + 4b^2+5b^2\) or, \(g(x) = (x – 2b)^2+5b^2\), Now, \((x2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get \((x2b)^2+5b^2≥5b^2\) Or, \(g(x)≥5b^2\) Option ECan you please further explain the need to factor the equation? I selected E because 9b^2 is the original equation. Any help would be super appreciated. Thanks Posted from my mobile devicehi syedazeem3will be glad to clarify your doubts. I would like to understand how you deduced option E simply by looking at \(9b^2\) in the original equation. If instead of \(5b^2\), option E had mentioned \(4b^2\), then would you still have chosen option E as correct?[/quote] My logic was, regardless of what is before 9b^2, 5b^2 would be smaller  the original equation would have to be larger than 5b^2. I would’ve probably picked E if it was 4b^2. Sent from my iPhone using GMAT Club Forum mobile app



Retired Moderator
Joined: 25 Feb 2013
Posts: 1167
Location: India
GPA: 3.82

Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
Show Tags
08 May 2018, 07:03
niks18 wrote: If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true
A.) \(g(x) < b\)
B.) \(g(x) < 0\)
C.) \(0 < g(x) < 4b\)
D.) \(g(x) ≥ 4b\)
E.) \(g(x) ≥ 5b^2\)
OE
\(g(x) = x^2 – 4bx + 4b^2+5b^2\)
or, \(g(x) = (x – 2b)^2+5b^2\),
Now, \((x2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get
\((x2b)^2+5b^2≥5b^2\)
Or, \(g(x)≥5b^2\)
Option E syedazeem3 wrote: My logic was, regardless of what is before 9b^2, 5b^2 would be smaller  the original equation would have to be larger than 5b^2. I would’ve probably picked E if it was 4b^2. Hi syedazeem3Note this is an algebraic equation where \(x\) is a variable and \(b\) is any constant and the question asks which relation for the equation ALWAYS HOLDS TRUE. By your logic if Option E had been \(8b^2\), then you would have selected it which is NOT ALWAYS TRUE Suppose \(x=2\) & \(b=1\) then \(8b^2=8*1^2=8\), but \(g(x) = x^2 – 4bx + 9b^2=2^24*2*1+9*1^2=5\) clearly here \(g(x)<8b^2\) in this case; but \(g(x)=5b^2\) So until you factories this quadratic equation you will not come to know that the function \(g(x)\) is always greater than or equal to \(5b^2\) for any value of \(x\) & \(b\)



Intern
Joined: 15 Jan 2016
Posts: 36

Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
Show Tags
08 May 2018, 18:46
niks18 wrote: niks18 wrote: If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true
A.) \(g(x) < b\)
B.) \(g(x) < 0\)
C.) \(0 < g(x) < 4b\)
D.) \(g(x) ≥ 4b\)
E.) \(g(x) ≥ 5b^2\)
OE
\(g(x) = x^2 – 4bx + 4b^2+5b^2\)
or, \(g(x) = (x – 2b)^2+5b^2\),
Now, \((x2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get
\((x2b)^2+5b^2≥5b^2\)
Or, \(g(x)≥5b^2\)
Option E syedazeem3 wrote: My logic was, regardless of what is before 9b^2, 5b^2 would be smaller  the original equation would have to be larger than 5b^2. I would’ve probably picked E if it was 4b^2. Hi syedazeem3Note this is an algebraic equation where \(x\) is a variable and \(b\) is any constant and the question asks which relation for the equation ALWAYS HOLDS TRUE. By your logic if Option E had been \(8b^2\), then you would have selected it which is NOT ALWAYS TRUE Suppose \(x=2\) & \(b=1\) then \(8b^2=8*1^2=8\), but \(g(x) = x^2 – 4bx + 9b^2=2^24*2*1+9*1^2=5\) clearly here \(g(x)<8b^2\) in this case; but \(g(x)=5b^2\) So until you factories this quadratic equation you will not come to know that the function \(g(x)\) is always greater than or equal to \(5b^2\) for any value of \(x\) & \(b\) Thanks for the help here niks18. Appreciate the insight  finally understood this. Posted from my mobile device




Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then
[#permalink]
08 May 2018, 18:46






