value of x

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jainrahul1985: Master | Next Rank: 500 Posts; Posts: 228; Joined: Sun Aug 17, 2008 8:08 am; Thanked: 4 times

by jainrahul1985 » Tue Oct 19, 2010 11:07 am

What is the value of |x|?

(1) |x^2 + 16| - 5 = 27

(2) x^2 = 8x - 16

OA to follow . Please explain the solution in detail

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Source: Beat The GMAT — Data Sufficiency | Tue Oct 19, 2010 11:07 am

shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Tue Oct 19, 2010 11:28 am

IMO B

2: x^2 = 8x - 16
=> x^2 - 8x + 16 = 0
=> ( x - 4 ) ^ 2 = 0
=> +(x-4) = 0 OR -(x-4) = 0
=> x = 4 OR -x = -4
|x| = 4 Sufficient

1. |x^2 + 16| - 5 = 27
=> |x^2 + 16| = 32
=> |x^2| = 16 = 4^2
=> |X| = 4 or |X| = -4 Not Sufficient
In this option if the ABSOLUTE symbol were not there, we would have got x = 4 OR -x = -4 (Or we can say |x| = 4) but as those are present it is always |x| which has 2 values.

Hope I m correct. Whats OA?

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uwhusky: Legendary Member; Posts: 1172; Joined: Wed Apr 28, 2010 6:20 pmFollowed by: 4 members; Thanked: 74 times

by uwhusky » Tue Oct 19, 2010 11:34 am

I'll go with D.

1) x^2 will always result in a positive number, and thus |x^2 + 16| is always positive even without the absolute value symbol. So there's only one solution for 1, x = 4 or -4. |x| therefore must be 4.

2) can be solved as (x - 4)(x - 4).

Yep.

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Tue Oct 19, 2010 11:53 am

uwhusky wrote: 1) x^2 will always result in a positive number, and thus |x^2 + 16| is always positive even without the absolute value symbol. So there's only one solution for 1, x = 4 or -4. |x| therefore must be 4.

I am not sure but just asking you...
|x^2 + 16| - 5 = 27
=> |x^2 + 16| = 32
=> |x^2| = 16 = 4^2
=> |X| = 4 or |X| = -4

Where does the logic fail here? If u agree with me up to 3rd step then take square root of both sides. Now sqrt(16) = +/- 4 right.
|sqrt(x^2)| = |+/- x| = |x|.

What is your opinion?

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uwhusky: Legendary Member; Posts: 1172; Joined: Wed Apr 28, 2010 6:20 pmFollowed by: 4 members; Thanked: 74 times

by uwhusky » Tue Oct 19, 2010 11:59 am

x^2 will always be a positive number or 0, agreed?

Therefore |x^2 + 16| is the same as x^2 + 16, and so the question can be rephrased as:

x^2 + 16 - 5 = 27

x^2 + 11 = 27

x^2 = 16

x = 4 or -4

|x| = 4.

I understand that there are certain standard methods in approaching a formula within absolute value, but given the information for this question, I think the absolute value is put in to confuse you.

Yep.

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Tue Oct 19, 2010 12:16 pm

uwhusky wrote:x^2 will always be a positive number or 0, agreed?

Agreed boss

Thanks

If the problem is Easy Respect it, if the problem is tough Attack it

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Oct 19, 2010 2:00 pm

jainrahul1985 wrote:What is the value of |x|?

(1) |x^2 + 16| - 5 = 27

Let's just worry about (1), since (2) has been well explained. My explanation basically boils down to the one already offered, I'm just breaking it down a bit differently for completeness.

We can simplify to:

|x^2 + 16| = 32

That means that either:

(x^2 + 16) = 32

x^2 = 16

x = -4 or +4

or

-(x^2 + 16) = 32

x^2 + 16 = -32

x^2 = -48

Well, since the GMAT is only concerned with real numbers, we can completely ignore the second solution (since root of -48 is an imaginary number).

So, we only need to worry about the first solution and we know that x is +/- 4.

Since the question asks for |x|, and since |4|=|-4|, (1) gives us a definite answer to the question and is also sufficient.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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jainrahul1985: Master | Next Rank: 500 Posts; Posts: 228; Joined: Sun Aug 17, 2008 8:08 am; Thanked: 4 times

by jainrahul1985 » Tue Oct 19, 2010 8:39 pm

OA D

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