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What is the value of integer x? (M2714) [#permalink]
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16 Jan 2013, 00:51
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What is the value of integer x? (1) 2x^2+9<9x (2) x+10=2x+8
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Re: What is the value of integer x? [#permalink]
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16 Jan 2013, 03:16



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Re: What is the value of integer x? [#permalink]
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17 Apr 2013, 14:46
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]
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17 Apr 2013, 14:58
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: \(x+10<0,x<10\) \(x10=2x+8\) \(x=6\) \(6\) 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]
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17 Apr 2013, 15:45
thnks that was subtle i frgot to check the range.....



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Re: M2714 What is the value of integer x? [#permalink]
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26 Apr 2013, 14:52
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: M2714 What is the value of integer x? [#permalink]
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26 Apr 2013, 14:57
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: M2714 What is the value of integer x? [#permalink]
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26 Apr 2013, 15:00
Is this approach completely NOT correct, or is this approach just missing the checkstep? What's the earliest stage in solving in which the exclusion can be spotted (by a notsopro fellow)?
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Re: M2714 What is the value of integer x? [#permalink]
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26 Apr 2013, 15:03
HumptyDumpty wrote: Is this approach completely NOT correct, or is this approach just missing the checkstep?
What's the earliest stage in solving in which the exclusion can be spotted (by a notsopro 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: M2714 What is the value of integer x? [#permalink]
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27 Apr 2013, 04:09
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:
\(2x^29x+9<0\) \(x^2\frac{9x}{2}+\frac{9}{2}<0\) \((x3)(x\frac{3}{2})<0\)
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 nonnegative, hence RHS must also be nonnegative: \(2x+8\geq0\) giving us \(x\geq4\). 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]
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10 Mar 2014, 05:47
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: \((x\frac{3}{2})(x3)<0\) > 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 nonnegative: \(2x+8\geq{0}\) > \(x\geq{4}\), 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: x24x94661.html#p731476 inequalitiestrick91482.htmleverythingislessthanzero108884.html?hilit=extreme#p868863xyplane71492.html?hilit=solving%20quadratic#p841486Hope it helps. this wud be realy easy question.. how did u change this in fraction..?2x2+9<9x.. (x3/2) (x3) ... ?? I can change integers..bt this fraction
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Re: What is the value of integer x? [#permalink]
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10 Mar 2014, 06:15
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: \((x\frac{3}{2})(x3)<0\) > 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 nonnegative: \(2x+8\geq{0}\) > \(x\geq{4}\), 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: x24x94661.html#p731476 inequalitiestrick91482.htmleverythingislessthanzero108884.html?hilit=extreme#p868863xyplane71492.html?hilit=solving%20quadratic#p841486Hope it helps. this wud be realy easy question.. how did u change this in fraction..?2x2+9<9x.. (x3/2) (x3) ... ?? I can change integers..bt this fraction If you cannot factor directly, then solve 2x^29x+9=0 to find the roots and factor that way. Factoring Quadratics: http://www.purplemath.com/modules/factquad.htmSolving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htmHope this helps.
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Re: What is the value of integer x? (M2714) [#permalink]
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11 Mar 2014, 10:01
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]
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12 Jun 2014, 11:38
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 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]
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12 Jun 2014, 11:45



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Re: What is the value of integer x ? I 2x^2 + 9 < 9x II x+10 [#permalink]
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12 Jun 2014, 11:59
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? (M2714) [#permalink]
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12 Jun 2014, 13:13
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? (M2714) [#permalink]
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03 Mar 2015, 16:54
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.htmHowever, 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 (2x3) and (x6) 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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