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If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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12 Aug 2018, 07:39
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If f(3x + 2) = 9x² + 12x  1, then f(k  1) = A) k²  2k  6 B) k²  2k  5 C) k²  2k  4 D) k²  2k + 1 E) k²  2k + 5 *Kudos for all correct solutions
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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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14 Aug 2018, 06:27
GMATPrepNow wrote: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
A) k²  2k  6 B) k²  2k  5 C) k²  2k  4 D) k²  2k + 1 E) k²  2k + 5
Given: f(3x + 2) = 9x² + 12x  1 First notice that 9x² + 12x  1 looks a lot like how (3x + 2)² looks when we expand an simplify it. Notice that (3x + 2)² = 9x² + 12x + 4This is VERY similar to 9x² + 12x  1 In fact, if we take 9x² + 12x + 4 and subtract 5, we get 9x² + 12x  1 That is: 9x² + 12x + 4  5 = 9x² + 12x  1 So, we can write: f( 3x + 2) = 9x² + 12x  1 = 9x² + 12x + 4  5= (3x + 2)²  5In other words, f( something) = (something)²  5So, for example, f( y) = y²  5And f( 7) = 7²  5Likewise, f( k  1) = (k  1)²  5= k²  2k + 1  5= k²  2k  4 Answer: C Cheers, Brent
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If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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12 Aug 2018, 07:58
GMATPrepNow wrote: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
A) k²  2k  6 B) k²  2k  5 C) k²  2k  4 D) k²  2k + 1 E) k²  2k + 5 Given, \(f(3x + 2)\) = \(9x^2 + 12x  1\)=\( 9x² + 12x +441=(3x+2)^2  5\) Or, \(f(k) = k^2  5\) , where k=3x+2 So, \(f(k1)=(k1)^25=k^22k+15=k^22k4\) Ans. (C)
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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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12 Aug 2018, 08:01
GMATPrepNow wrote: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
A) k²  2k  6 B) k²  2k  5 C) k²  2k  4 D) k²  2k + 1 E) k²  2k + 5
*Kudos for all correct solutions 3x+2 = k1 3x = k3 Put the value of 3x in 9x^2 + 12x  1 f(k1)=f(3x+2)= 9x^2 + 12x  1 f(k1)=f(3x+2)= 3x*3x + 4*3x 1 now put 3x = k3 f(k1)=f(3x+2)= (k3)(k3) + 4(k3) 1 f(k1)=f(3x+2)= k^2  2k  4
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If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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12 Aug 2018, 08:03
GMATPrepNow wrote: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
A) k²  2k  6 B) k²  2k  5 C) k²  2k  4 D) k²  2k + 1 E) k²  2k + 5
*Kudos for all correct solutions f(3x + 2) =\(9x^2 + 12x  1\) We see that \((3x+2)^2\) = \(9x^2 + 12x + 4\) , if we subtract 5 from this, we get the desired value. So, f(3x + 2) = \((3x+2)^2  5\) = \(9x^2 + 12x  1\) f(k1) = \((k1)^25\) = \(k^2  2k +1  5\) = \(k^2 2k  4\) Answer C.
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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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12 Aug 2018, 09:38
3x+2 = k1 x= k3/3 9(k3/3)^2 +12(k3/3) 1 K^2 + 9 6k +4k 12 1 K^2 2k 4
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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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12 Aug 2018, 10:24
If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
A) k²  2k  6 B) k²  2k  5 C) k²  2k  4 D) k²  2k + 1 E) k²  2k + 5
Given, f(3x+2)^2 = 9x^2+12x−1 (3x+2)^2 =9x^2+12x+4 F(x)=x^25 So f(k1)=k²  2k  4
B option
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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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12 Aug 2018, 11:23
PKN wrote: GMATPrepNow wrote: If f(3x + 2) = 9x² + 12x  1[/color], then f(k  1) =
A) k²  2k  6 B) k²  2k  5 C) k²  2k  4 D) k²  2k + 1 E) k²  2k + 5 Alternate approach: Let k=1, with the result that f(k1) = f(0). The question stem becomes: If f(3x + 2) = 9x² + 12x  1, what is the value of f(0)? Since f(3x+2) = f(0) when x=2/3, plug x=2/3 into the expression in blue: \(9(\frac{2}{3})^2 + 12(\frac{2}{3})  1 = 4  8  1 = 5\) Now plug k=1 into the answer choices to see which yields the target value of 5. Only C works: \(k^2  2k  4 = 1  2  4 = 5\)
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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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12 Aug 2018, 12:41
HKD1710 wrote: GMATPrepNow wrote: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
A) k²  2k  6 B) k²  2k  5 C) k²  2k  4 D) k²  2k + 1 E) k²  2k + 5
*Kudos for all correct solutions 3x+2 = k1 3x = k3 Put the value of 3x in 9x^2 + 12x  1 f(k1)=f(3x+2)= 9x^2 + 12x  1 f(k1)=f(3x+2)= 3x*3x + 4*3x 1 now put 3x = k3 f(k1)=f(3x+2)= (k3)(k3) + 4(k3) 1 f(k1)=f(3x+2)= k^2  2k  4 awesome approach



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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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13 Aug 2018, 01:46
PKN wrote: GMATPrepNow wrote: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
A) k²  2k  6 B) k²  2k  5 C) k²  2k  4 D) k²  2k + 1 E) k²  2k + 5 Given, \(f(3x + 2)\) = \(9x^2 + 12x  1\)=\( 9x² + 12x +441=(3x+2)^2  5\) Or, \(f(k) = k^2  5\) , where k=3x+2 So, \(f(k1)=(k1)^25=k^22k+15=k^22k4\) Ans. (C) In the any algebraic approach from above, why equating (3x + 2)=k1?? I do not understand this. Anyone can help? Thanks in advance



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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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13 Aug 2018, 02:01
GMATPrepNow wrote: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
A) k²  2k  6 B) k²  2k  5 C) k²  2k  4 D) k²  2k + 1 E) k²  2k + 5
*Kudos for all correct solutions OA:CLet \(3x+2 = y\) \(3x =y2\) \(x= \frac{y2}{3}\) \(f(3x + 2) = 9x² + 12x  1\) \(f(y) = 9 (\frac{y2}{3})^2 +12(\frac{y2}{3}) 1\) \(f(y) = (y2)^2 +4(y2) 1=y^2+42*2y + 4y 8 1 =y^25\) \(f(y)=y^25\) \(f(k  1) =(k1)^25 =k^2+12k5 =k^22k4\)



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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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14 Aug 2018, 08:33
GMATPrepNow wrote: In other words, f(something) = (something)²  5 So, for example, f(y) = y²  5 And f(7) = 7²  5
Likewise, f(k  1) = (k  1)²  5 = k²  2k + 1  5 = k²  2k  4
Answer: C
Cheers, Brent
Dear Brent, Very enlightening answer. Your way above answered my question in simple terms. Thanks



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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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14 Aug 2018, 09:08
Thanks Mo2men! Cheers Brent Posted from my mobile device
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Re: If f(3x + 2) = 9x² + 12x  1, then f(k  1) =
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