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If |(x – 3)^2 + 2| < |x – 7| , which of the following expresses the al

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Magoosh GMAT Instructor
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If |(x – 3)^2 + 2| < |x – 7| , which of the following expresses the al  [#permalink]

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New post 26 Nov 2016, 12:20
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A
B
C
D
E

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Question Stats:

71% (02:37) correct 29% (02:50) wrong based on 324 sessions

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If \(|(x – 3)^2 + 2| < |x – 7|\) , which of the following expresses the allowable range for x?

(A) 1 < x < 4

(B) 1 < x < 7

(C) – 1 < x < 4 and 7 < x

(D) x < – 1 and 4 < x < 7

(E) – 7 < x < 4 and 7 < x


Absolute value inequalities are a rare and tricky topic on the GMAT. For a detailed discussion of this question type, as well as the OE for this particular question, see:
Absolute Value Inequalities

Mike :-)

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If |(x – 3)^2 + 2| < |x – 7| , which of the following expresses the al  [#permalink]

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New post 27 Nov 2016, 00:03
6
Hi,

I always go for plugging in for these kind of questions. Its fast and quite reliable.

Lets look at one option at at time:-

A. 1<x<4: Lets take X= 2 and 3, the equation is satisfied. Hold on to it.

B. 1<x<7: We have already tested the equation for x=2 and 3. Now, we can check it with X=5, which gives the result as false. So, discard this option.

C. -1<x<4 and x<7: In the above statement, we have already tested for X=5 or we can say that for the second part of this statement x<7. So, this statement is false. However, just to be sure, putting x=0 will give false. So, discard.

D. X<-1 and 4<x<7: As done earlier, the statement has proven to be false for X=5. So, this statement can also be discarded.

E. -7<x<4 and x<7: Already tested for X=0, 2 and 5. So discard.

Thus the correct answer is A.
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If |(x – 3)^2 + 2| < |x – 7| , which of the following expresses the al  [#permalink]

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New post 26 Nov 2016, 19:41
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mikemcgarry wrote:
If \(|(x – 3)^2 + 2| < |x – 7|\) , which of the following expresses the allowable range for x?

(A) 1 < x < 4

(B) 1 < x < 7

(C) – 1 < x < 4 and 7 < x

(D) x < – 1 and 4 < x < 7

(E) – 7 < x < 4 and 7 < x


Note that \((x-3)^2+2 \geq 2 >0 \quad\forall x \implies |(x – 3)^2 + 2| = (x – 3)^2 + 2\)

If \(x \geq 7 \implies (x – 3)^2 + 2 < x – 7 \implies x^2 -6x +9+2<x-7 \implies x^2 -7x +18 <0\)
We have \(x^2 -7x+18 = x^2 -2 \times x \times \frac{7}{2} +(\frac{7}{2})^2 +\frac{23}{4} = (x-\frac{7}{2})^2 +\frac{23}{4} >0\)
So in this case, there is no \(x\) satisfied the equation.

If \(x <7 \implies (x – 3)^2 + 2 < 7-x \implies x^2 -6x +9+2<7-x \implies x^2 -5x +4 <0\)
\(\implies (x-1)(x-4)<0 \implies 1<x<4\)

The answer is A
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Re: If |(x – 3)^2 + 2| < |x – 7| , which of the following expresses the al  [#permalink]

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New post 17 Jun 2018, 19:50
I looked at the options and substituted 6 and saw anything greater than 6 was also not going to work and was able to eliminate B,C,D,E in one shot.
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Re: If |(x – 3)^2 + 2| < |x – 7| , which of the following expresses the al  [#permalink]

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New post 09 Jul 2018, 05:04
broall wrote:
mikemcgarry wrote:
If \(|(x – 3)^2 + 2| < |x – 7|\) , which of the following expresses the allowable range for x?

(A) 1 < x < 4

(B) 1 < x < 7

(C) – 1 < x < 4 and 7 < x

(D) x < – 1 and 4 < x < 7

(E) – 7 < x < 4 and 7 < x


Note that \((x-3)^2+2 \geq 2 >0 \quad\forall x \implies |(x – 3)^2 + 2| = (x – 3)^2 + 2\)

If \(x \geq 7 \implies (x – 3)^2 + 2 < x – 7 \implies x^2 -6x +9+2<x-7 \implies x^2 -7x +18 <0\)
We have \(x^2 -7x+18 = x^2 -2 \times x \times \frac{7}{2} +(\frac{7}{2})^2 +\frac{23}{4} = (x-\frac{7}{2})^2 +\frac{23}{4} >0\)
So in this case, there is no \(x\) satisfied the equation.

If \(x <7 \implies (x – 3)^2 + 2 < 7-x \implies x^2 -6x +9+2<7-x \implies x^2 -5x +4 <0\)
\(\implies (x-1)(x-4)<0 \implies 1<x<4\)

The answer is A




Yes the answer may seem correct but I don't understand how (x−1)(x−4)<0⟹1<x<4⟹(x−1)(x−4)<0⟹1<x<4
Shouldn't it be x< 1 and x<4 and the correct answer just x<4 It looks like a subset.

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Re: If |(x – 3)^2 + 2| < |x – 7| , which of the following expresses the al  [#permalink]

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New post 09 Jul 2018, 05:41
Based on Answer Choices, try x=5
Inequality not true
Eliminate B, C, D, E
Answer A
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Re: If |(x – 3)^2 + 2| < |x – 7| , which of the following expresses the al &nbs [#permalink] 09 Jul 2018, 05:41
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