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What is the value of integer x? (M27-14)

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What is the value of integer x? (M27-14) [#permalink] New post 15 Jan 2013, 23:51
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What is the value of integer x?

(1) 2x^2+9<9x
(2) |x+10|=2x+8
[Reveal] Spoiler: OA
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Re: What is the value of integer x? [#permalink] New post 16 Jan 2013, 02:16
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LM wrote:
What is the value of integer x?

(1) 2x^2+9<9x
(2) |x+10|=2x+8


What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.

Check this for more on solving inequalities like the one in the first statement:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.
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Re: What is the value of integer x? [#permalink] New post 17 Apr 2013, 13:46
I understand that statement 1 is sufficient....bt
Statement 2, i tried solving and gt two different answers....

lx+10l = 2x+8
when x>0
x+10 =2x+8
x=2
Another case when x<0
-(x+10) =2x+8
x=-6

Am i wrong while solving the mod....

Help me with statement 2

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Re: What is the value of integer x? [#permalink] New post 17 Apr 2013, 13:58
Hi Archit143,

Lets see statement 2: (2) |x+10|=2x+8

We have to analyze the two cases so
1)Range: \(x+10>0, x>-10\)
\(x+10=2x+8\)
\(x=2\)
\(-2\) is in the range we are analyzing (>-10) so -2 is a solution

2)Range: \(x+10<0,x<-10\)
\(-x-10=2x+8\)
\(x=-6\)
\(-6\) is OUT of the range we are considering (<-10) so is NOT a possible solution

Let me know if it's clear
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Re: What is the value of integer x? [#permalink] New post 17 Apr 2013, 14:45
thnks that was subtle i frgot to check the range.....
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Re: M27-14 What is the value of integer x? [#permalink] New post 26 Apr 2013, 13:52
Expert's post
HumptyDumpty wrote:
My way:
standard procedure:
1st: \(x+10=2x+8\)
\(x=2\)
OR
2nd:\(-(x+10)=2x+8\)
\(x=-6\)
the condition \(x=i\) is met:
Insufficient.

This approach is NOT correct.

Always be sure to plug back the solutions for equations with modulus sign
the values that you have found are x = 2 and x = -6
ONLY x=2 satisfies the given condition
For X = -6

LHS = |x+10| = 4
RHS = 2x + 8 = -4
SO x = -6 is NO GOOD

This is happening because you have found the roots of
\((x+10)^2\) = \((2x+8)^2\)
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Re: M27-14 What is the value of integer x? [#permalink] New post 26 Apr 2013, 13:57
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This could ALSO be solved YOUR way but you have missed the definition of the modulus sign
|x|= x WHEN x >or = 0
|x| = -x WHEN x<0

Now lets come back to the problem,

|x+10| = 2x+8

SO x+10 = 2x+8 when x+10> 0
So it gives x=2
Now lets check: is 2+10 > 0 YES. So this is GOOD
Next
-(x+10) = 2x+8 when x+10<0
it gives x=-6
Lets check: is -6+10 < 0 NO. So this is what you call an extraneous root. Does no good.
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Re: M27-14 What is the value of integer x? [#permalink] New post 26 Apr 2013, 14:00
Is this approach completely NOT correct,
or
is this approach just missing the check-step?

What's the earliest stage in solving in which the exclusion can be spotted (by a not-so-pro fellow)?
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Re: M27-14 What is the value of integer x? [#permalink] New post 26 Apr 2013, 14:03
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HumptyDumpty wrote:
Is this approach completely NOT correct,
or
is this approach just missing the check-step?

What's the earliest stage in solving in which the exclusion can be spotted (by a not-so-pro fellow)?


Well you just need to know the definition of modulus

sure |x| = 8 will give you 8 and -8 as solution but that will not hold true when the RHS also has a variable

when both sides consist variable go by the definition.

You have only taken a part of the definition. You need to look at the entire scene as explained in the second post.
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Re: M27-14 What is the value of integer x? [#permalink] New post 27 Apr 2013, 03:09
Expert's post
HumptyDumpty wrote:
What is the value of integer x?

1) \(2x^2+9<9x\)
2) \(|x+10|=2x+8\)


1) is clear: compute the quadratic expression, consider the scope of values and the condition x=i:

\(2x^2-9x+9<0\)
\(x^2-\frac{9x}{2}+\frac{9}{2}<0\)
\((x-3)(x-\frac{3}{2})<0\)

the sums in parentheses must have opposite signs, so:

\(\frac{3}{2}<x<3\)

consider the condition \(x=i\):

\(x=2\)
Sufficient.

2) is not clear:

\(|x+10|=2x+8\)

My way:
standard procedure:
1st: \(x+10=2x+8\)
\(x=2\)
OR
2nd:\(-(x+10)=2x+8\)
\(x=-6\)
the condition \(x=i\) is met:
Insufficient.

The original explanation:
(2) \(|x+10|=2x+8\). The left hand side (LHS) is an absolute value, which is always non-negative, hence RHS must also be non-negative: \(2x+8\geq0\) giving us \(x\geq-4\). Now, for this range \(x+10\) is positive, hence\(|x+10|=x+10\). So, \(|x+10|=2x+8\) can be written as \(x+10=2x+8\), solving for \(x\) gives \(x=2\).
Sufficient.


Noting that \(2x+8\geq0\) excludes the negative value and leaves off only one value. But what the heck is the mechanics behind this problem that makes the the good old way of solving inequalities insufficiently precise here?


Merging similar topics.

The way you call "standard procedure" is not complete.

When expanding \(|x+10|=2x+8\) you should consider x<-10 range and x>=-10 range:

\(x<-10\) --> \(-(x+10)=2x+8\) --> \(x=-6\). Discard this solution since it's not in the range \(x<-10\).
\(x\geq{-10}\) --> \(x+10=2x+8\) --> \(x=2\).

Hope it's clear.
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Re: What is the value of integer x? [#permalink] New post 10 Mar 2014, 04:47
Bunuel wrote:
LM wrote:
What is the value of integer x?

(1) 2x^2+9<9x
(2) |x+10|=2x+8


What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.

Check this for more on solving inequalities like the one in the first statement:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.


this wud be realy easy question..

how did u change this in fraction..?2x2+9<9x.. (x-3/2) (x-3) ... ?? I can change integers..bt this fraction :(
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Re: What is the value of integer x? [#permalink] New post 10 Mar 2014, 05:15
Expert's post
sanjoo wrote:
Bunuel wrote:
LM wrote:
What is the value of integer x?

(1) 2x^2+9<9x
(2) |x+10|=2x+8


What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.

Check this for more on solving inequalities like the one in the first statement:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.


this wud be realy easy question..

how did u change this in fraction..?2x2+9<9x.. (x-3/2) (x-3) ... ?? I can change integers..bt this fraction :(


If you cannot factor directly, then solve 2x^2-9x+9=0 to find the roots and factor that way.

Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm

Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm

Hope this helps.
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Re: What is the value of integer x? (M27-14) [#permalink] New post 11 Mar 2014, 09:01
Is this a type of question one can expect on the GMAT ?

The factoring of the quadratic isn't "clean", as opposed to one that can be solved using integers.
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What is the value of integer x ? I- 2x^2 + 9 < 9x II- |x+10 [#permalink] New post 12 Jun 2014, 10:38
What is the value of integer x ?

I- 2x^2 + 9 < 9x
II- |x+10| = 2x+8

Bunuel or Karishma..Please explain how to approach statement II
I got the answer
[Reveal] Spoiler:
A
with my standard approach i.e opening the absolute bracket in two different cases - x less than and greater than 0
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Re: What is the value of integer x ? I- 2x^2 + 9 < 9x II- |x+10 [#permalink] New post 12 Jun 2014, 10:45
Expert's post
JusTLucK04 wrote:
What is the value of integer x ?

I- 2x^2 + 9 < 9x
II- |x+10| = 2x+8

Bunuel or Karishma..Please explain how to approach statement II
I got the answer
[Reveal] Spoiler:
A
with my standard approach i.e opening the absolute bracket in two different cases - x less than and greater than 0


Merging similar topics. Please refer to the discussion above.
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Re: What is the value of integer x ? I- 2x^2 + 9 < 9x II- |x+10 [#permalink] New post 12 Jun 2014, 10:59
What is the value of integer x ?

I- 2x^2 + 9 < 9x
II- |x+10| = 2x+8

I took some time but solved it my way.

IMO D.

Stmt I. we can make equation \(2x^2 -9x + 9 < 0\)

solve for X = 1.5, 3 -- 2 is the only integer in between satisfies the relation.

Sufficient.

II. Square both side.
\((|x+10|) ^2=(2x+8)^2\)

\(x^2 + 4x - 12 = 0\)

solve for X = 2, -6 -- substitute values back to original equation, only 2 satisfies the relation.

Sufficient.
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Re: What is the value of integer x? (M27-14) [#permalink] New post 12 Jun 2014, 12:13
Sad..I carelessly took X>0 instead of X>10.. and hence both the answers for statement II fitted in :(
Nice question
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What is the value of integer x? (M27-14) [#permalink] New post 03 Mar 2015, 15:54
Dear Bunuel, I always have problem with factoring Quadratics so, I tried to use the way that I found in the link 1 that

you add.Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm

However, I did not get the right answer. that is how I answered the question:

The coefficients are a = 2, b = –9, and c = 9, so ac = 18. Since 18 is positive, I need two factors that are either both

positive or else both negative . Since –9 is negative, I need the factors both to be negative. The pairs of factors for 18

are -3 and -6. Since -6+(-3) = –9, I will use –6 and –3 so the roots are (2x-3) and (x-6) which mean x=3/2 , x=6

so based on statement 1 x=6

based on statemebt 2 x=2 what I missed here ? What is the wrong in my way?
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What is the value of integer x? (M27-14)   [#permalink] 03 Mar 2015, 15:54
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