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805+ Level|   Algebra|   Exponents|      
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Bunuel
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If 2^(2x+4) – 17*2^(x+1) = –4, then what is the sum of all possible va 21 Apr 2020, 08:27
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95% (hard)

Question Stats:

43% (02:36) correct 57% (02:43) wrong based on 123 sessions

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If \(2^{(2x+4)} – 17*2^{(x+1)} = –4\), then what is the sum of all possible values of x ?

A. -2
B. -1
C. 0
D. 1
E. 2

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A
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If 2^(2x+4) – 17*2^(x+1) = –4, then what is the sum of all possible va 21 Apr 2020, 08:51
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Nice question Bunuel

Substituting \(Y = 2^(^x^+^1^)\), we have:
\(4Y^2-17Y + 4 = 0\)

The solutions are Y = 4 or Y =1/4
Hence x+1 = 2 or x+1 = -2
Hence x = 1 or x = -3

Summing both solutions we get 1-3 = -2
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If 2^(2x+4) – 17*2^(x+1) = –4, then what is the sum of all possible va 24 Apr 2020, 00:48
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Nice question Bunuel

Substituting \(Y = 2^(^x^+^1^)\), we have:
\(4Y^2-17Y + 4 = 0\)

The solutions are Y = 4 or Y =1/4
Hence x+1 = 2 or x+1 = -2
Hence x = 1 or x = -3

Summing both solutions we get 1-3 = -2
Answer (A)
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Hello !! I couldn't understand the highlighted part.

Here is my doubt.

2^(2x+4) – 17*2^(x+1) = –4
4^(x+2) – 17*2^(x+1) = –4

How could you 4^(x+2) = 4Y^2

Thanks in advance!!! :)
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GMAT Focus 1: 735 Q90 V89 DI81
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If 2^(2x+4) – 17*2^(x+1) = –4, then what is the sum of all possible va 25 Apr 2020, 00:57
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Bunuel
If \(2^{(2x+4)} – 17*2^{(x+1)} = –4\), then what is the sum of all possible values of x ?

A. -2
B. -1
C. 0
D. 1
E. 2

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Show more


\(2^{(2x+4)} – 17*2^{(x+1)} = –4\)....
\(2^{(2x+2+2)} – 17*2^{(x+1)} = –4\).....
\(2^{2(x+1)+2} – 17*2^{(x+1)} = –4\)....
\(2^{2(x+1)}*2^2 – 17*2^{(x+1)} = –4\)..
\((2^{(x+1)})^2*2^2 – 17*2^{(x+1)} = –4\)..

Now, to simplify the term, we can take a=\(2^{(x+1)}\)
\((a)^2*4 – 17*a = –4\)..
So, \(4a^2-17a+4=0.......(4a-1)(a-4)=0\)

Thus a=\(2^{(x+1)}=4=2^2\).....\(x+1=2......x=1\)
Or a=\(2^{(x+1)}=\frac{1}{4}=\frac{1}{2^2}=2^{-2}\).....\(x+1=-2......x=-3\)

Sum of values = \(1+(-3)=-2\)

A
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If 2^(2x+4) – 17*2^(x+1) = –4, then what is the sum of all possible va 14 Apr 2024, 07:56
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Asked: If \(2^{(2x+4)} – 17*2^{(x+1)} = –4\), then what is the sum of all possible values of x ?

\(2^{(2x+4)} – 17*2^{(x+1)} = –4\)
\(16*2^{2x} – 34*2^x = –4\)

Let 2^x = t

\(16t^2 - 34t = -4\)
\(8t^2 - 17t + 2 = 0\)
\(8t^2 - 16t - t + 2 = 0\)
(8t - 1)(t-2) = 0
t = 1/8 or 2

x = -3 or 1

The sum of all possible values of x = -3 + 1 = -2

IMO A
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