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16 Sep 2014, 01:27
Official Solution: (1) \(2x^2+9 \lt 9x\). Factor quadratics: \((x\frac{3}{2})(x3) \lt 0\). The roots are \(\frac{3}{2}\) and 3, the "\(\lt\)" sign indicates that the solution lies between the roots: \(1.5 \lt x \lt 3\). Since the only integer in this range is 2, then \(x=2\). Sufficient. (2) \(x+10=2x+8\). The left hand side (LHS) is an absolute value, which is always nonnegative, hence RHS must also be nonnegative: \(2x+8 \ge 0\) giving us \(x \ge 4\). 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. Answer: D
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18 Sep 2014, 18:25
Bunuel wrote: Official Solution:
(1) \(2x^2+9 \lt 9x\). Factor quadratics: \((x\frac{3}{2})(x3) \lt 0\). The roots are \(\frac{3}{2}\) and 3, the "\(\lt\)" sign indicates that the solution lies between the roots: \(1.5 \lt x \lt 3\). Since the only integer in this range is 2, then \(x=2\). Sufficient. (2) \(x+10=2x+8\). The left hand side (LHS) is an absolute value, which is always nonnegative, hence RHS must also be nonnegative: \(2x+8 \ge 0\) giving us \(x \ge 4\). 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.
Answer: D Hi Bunuel, Could you please explain the statement A of the solution. How is the root (x3/2)? The roots that I am getting are (x3) and (2x3). Please could you help me out with this. Thanks,



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19 Sep 2014, 01:30
msbcapitalist wrote: Bunuel wrote: Official Solution:
(1) \(2x^2+9 \lt 9x\). Factor quadratics: \((x\frac{3}{2})(x3) \lt 0\). The roots are \(\frac{3}{2}\) and 3, the "\(\lt\)" sign indicates that the solution lies between the roots: \(1.5 \lt x \lt 3\). Since the only integer in this range is 2, then \(x=2\). Sufficient. (2) \(x+10=2x+8\). The left hand side (LHS) is an absolute value, which is always nonnegative, hence RHS must also be nonnegative: \(2x+8 \ge 0\) giving us \(x \ge 4\). 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.
Answer: D Hi Bunuel, Could you please explain the statement A of the solution. How is the root (x3/2)? The roots that I am getting are (x3) and (2x3). Please could you help me out with this. Thanks, Yes, \(2x^2  9x +9<0\) can be factored as \((x  3)(2x  3)<0\) but notice that we can factor 2 out of (2x  3) and reduce by it to get \((x3)(x\frac{3}{2}) \lt 0\). Theory on InequalitiesSolving Quadratic Inequalities  Graphic Approach: solvingquadraticinequalitiesgraphicapproach170528.htmlInequality tips: tipsandhintsforspecificquanttopicswithexamples172096.html#p1379270inequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.htmleverythingislessthanzero108884.htmlgraphicapproachtoproblemswithinequalities68037.html
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20 Sep 2014, 16:14
Bunuel the "\lt" sign indicates that the solution lies between the roots: 1.5 \lt x \lt 3. Can you please elaborate on this statement ? How did we reach that conclusion? Thank you



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22 Sep 2014, 15:02
Bunuel wrote: ShravanHemchand wrote: Bunuel the "\lt" sign indicates that the solution lies between the roots: 1.5 \lt x \lt 3. Can you please elaborate on this statement ? How did we reach that conclusion? Thank you Check here: solvingquadraticinequalitiesgraphicapproach170528.htmlBunuel Can you please explain statement 'B'? The part I missed was that why didn't you consider the negative of the LHS, and just the +ve?



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23 Sep 2014, 01:11
p2bhokie wrote: Bunuel wrote: ShravanHemchand wrote: Bunuel the "\lt" sign indicates that the solution lies between the roots: 1.5 \lt x \lt 3. Can you please elaborate on this statement ? How did we reach that conclusion? Thank you Check here: solvingquadraticinequalitiesgraphicapproach170528.htmlBunuel Can you please explain statement 'B'? The part I missed was that why didn't you consider the negative of the LHS, and just the +ve? Absolute value on the left hand side (x + 10) cannot be negative: \(some \ expression\geq{0}\). Thus 2x + 8, on the right hand side must also be nonnegative, which means that \(x \ge 4\). Now, if \(x \ge 4\), then x + 10 will be positive, and we know that if x>0, then x = x, so x + 10 = x + 10. Similar questions to practice: isx157102.htmlwhatisx126874.htmlism01mm2m99913.htmlwhatisthevalueofx161134.htmlwhatisthevalueofx169753.htmlwhatisthevalueofy103925.htmlwhatisthevalueofy127976.htmlisx01x34x32x32x127978.htmlifx0isx11x212x1x130402.htmlisx01x34x32x12x1can100357.htmlwhatisthevalueofintegerxm27145839.htmlthediscreetcharmoftheds12696240.html#p1039650Hope this helps.
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02 Nov 2014, 07:13
Can you please tell, why are we taking Absolute value as positive? I have seen in majority of the questions that when we solve Absolute value equation, it gives two solutions. one negative and other positive For Example: theareaboundedbythecurvesxy1andxy1is93103.htmlcan you please explain the difference? where we have take both values and where we have to take only positive value



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02 Nov 2014, 07:52
awal_786@hotmail.com wrote: Can you please tell, why are we taking Absolute value as positive? I have seen in majority of the questions that when we solve Absolute value equation, it gives two solutions. one negative and other positive For Example: theareaboundedbythecurvesxy1andxy1is93103.htmlcan you please explain the difference? where we have take both values and where we have to take only positive value The absolute value of a number is the value of a number without regard to its sign. For example, 3 = 3; 12 = 12; 1.3=1.3... Another way to understand absolute value is as the distance from zero. For example, x is the distance between x and 0 on a number line. From that comes the most important property of an absolute value: since the distance cannot be negative, an absolute value expression is ALWAYS more than or equal to zero.
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25 Oct 2015, 13:39
BunuelOption B can be solved as X 1. x+10=(2x+8) .....here x= 2/3 2. x+10=2x+8....here x=2 as x is intiger...x=2..sufficient Is this approach correct?



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10 Sep 2016, 05:13
(1) 2x2+9<9x2x2+9<9x. Factor quadratics: (x−32)(x−3)<0(x−32)(x−3)<0. The roots are 3232 and 3, the "<<" sign indicates that the solution lies between the roots: 1.5<x<31.5<x<3. Since the only integer in this range is 2, then x=2x=2. Sufficient.
bunuel could u plz explain how are u getting the roots in st1??? how to approach quadratic equation when there is an integer before x^2?



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Bunuel wrote: Official Solution:
(1) \(2x^2+9 \lt 9x\). Factor quadratics: \((x\frac{3}{2})(x3) \lt 0\). The roots are \(\frac{3}{2}\) and 3, the "\(\lt\)" sign indicates that the solution lies between the roots: \(1.5 \lt x \lt 3\). Since the only integer in this range is 2, then \(x=2\). Sufficient. (2) \(x+10=2x+8\). The left hand side (LHS) is an absolute value, which is always nonnegative, hence RHS must also be nonnegative: \(2x+8 \ge 0\) giving us \(x \ge 4\). 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.
Answer: D if \((x\frac{3}{2})(x3) \lt 0\) then wont x3/2 < 0 therefore x<3/2 and x3<0 therefore x<3 then if x is less than 3 and 3/2 at the same time, how does one arrive at x<2 Am I doing this correct?



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04 Nov 2016, 08:30
vdhaval wrote: Bunuel wrote: Official Solution:
(1) \(2x^2+9 \lt 9x\). Factor quadratics: \((x\frac{3}{2})(x3) \lt 0\). The roots are \(\frac{3}{2}\) and 3, the "\(\lt\)" sign indicates that the solution lies between the roots: \(1.5 \lt x \lt 3\). Since the only integer in this range is 2, then \(x=2\). Sufficient. (2) \(x+10=2x+8\). The left hand side (LHS) is an absolute value, which is always nonnegative, hence RHS must also be nonnegative: \(2x+8 \ge 0\) giving us \(x \ge 4\). 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.
Answer: D if \((x\frac{3}{2})(x3) \lt 0\) then wont x3/2 < 0 therefore x<3/2 and x3<0 therefore x<3 then if x is less than 3 and 3/2 at the same time, how does one arrive at x<2 Am I doing this correct? The correct range is \(1.5 \lt x \lt 3\), so no you are not doing it right. x < 3 AND x < 1.5 does not make sense. What does it even mean? Can x be 2 since x < 3 or not since x < 1.5. Anyway, please check the links below to study how to derive the ranges for inequalities: Inequalities Made Easy!Solving Quadratic Inequalities  Graphic ApproachInequality tipsWavy Line Method Application  Complex Algebraic InequalitiesDS Inequalities Problems PS Inequalities Problems 700+ Inequalities problemsinequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.htmleverythingislessthanzero108884.htmlgraphicapproachtoproblemswithinequalities68037.htmlHope this helps.
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16 Dec 2016, 12:20
BunuelHis question still stands (the part that's also confusing me too). If (x3/2)(x3) < 0, then algebraically (x<3) and (x<3/2). To answer your question, one could think x<3/2 is more limiting than x<3. Per the answer derived from your solution (x3/2)(x3) < 0 when solving for x, the question we ask is why won't the sign change from x<3/2 to x>3/2? There has to be a negative division in order to change the sign in my knowledge. Thank you for your time!



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Hi Bunuel,
Thanks for the factoring link...I was having problems factoring quadratics where a was greater than 1.
Anyhow...with my newfound knowledge I attempted this particular question and still came unstuck at one final stage...
Yes I understand that (x3) (2x3)<0 ... I then did the following:
x3<0 ... so x<3 2x3<0 ... so 2x<3...x<1.5
How can x be less than 3 and less than 1.5 and thus be an integer 2?
I clearly went wrong somewhere but I don't know where.
Thanks,
Tosin



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15 Feb 2018, 23:22
ttaiwo wrote: Hi Bunuel,
Thanks for the factoring link...I was having problems factoring quadratics where a was greater than 1.
Anyhow...with my newfound knowledge I attempted this particular question and still came unstuck at one final stage...
Yes I understand that (x3) (2x3)<0 ... I then did the following:
x3<0 ... so x<3 2x3<0 ... so 2x<3...x<1.5
How can x be less than 3 and less than 1.5 and thus be an integer 2?
I clearly went wrong somewhere but I don't know where.
Thanks,
Tosin (x  3)(2x  3) < 0 to be true the multiples, x  3 and 2x  3 must have different signs. x  3 > 0 and 2x  3 < 0 > x > 3 and x < 3/2. Both cannot be simultaneously true. So, this case is out. x  3 < 0 and 2x  3 > 0 > x < 3 and x > 3/2. So, 3/2 < x < 3.
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Re: M2714
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16 Feb 2018, 01:12
Bunuel wrote: ttaiwo wrote: Hi Bunuel,
Thanks for the factoring link...I was having problems factoring quadratics where a was greater than 1.
Anyhow...with my newfound knowledge I attempted this particular question and still came unstuck at one final stage...
Yes I understand that (x3) (2x3)<0 ... I then did the following:
x3<0 ... so x<3 2x3<0 ... so 2x<3...x<1.5
How can x be less than 3 and less than 1.5 and thus be an integer 2?
I clearly went wrong somewhere but I don't know where.
Thanks,
Tosin (x  3)(2x  3) < 0 to be true the multiples, x  3 and 2x  3 must have different signs. x  3 > 0 and 2x  3 < 0 > x > 3 and x < 3/2. Both cannot be simultaneously true. So, this case is out. x  3 < 0 and 2x  3 > 0 > x < 3 and x > 3/2. So, 3/2 < x < 3. Thanks bro. Was cracking my brain all day yesterday on that one.



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30 May 2018, 11:51
Bunuel wrote: Official Solution:
(1) \(2x^2+9 \lt 9x\). Factor quadratics: \((x\frac{3}{2})(x3) \lt 0\). The roots are \(\frac{3}{2}\) and 3, the "\(\lt\)" sign indicates that the solution lies between the roots: \(1.5 \lt x \lt 3\). Since the only integer in this range is 2, then \(x=2\). Sufficient. (2) \(x+10=2x+8\). The left hand side (LHS) is an absolute value, which is always nonnegative, hence RHS must also be nonnegative: \(2x+8 \ge 0\) giving us \(x \ge 4\). 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.
Answer: D Hi Bunuel , I understood statement A , however , I'm not able to understand statement 2 , in this , X can hold values from 4<x<2 , then X can be 1 also ?? Please guide ,where I'm going wrong. I selected E , while taking the gmat club Quant Test , as I got the same answer as yours , in statement 1, however I missed out X is an integer , in that case statement 1 holds true , in statement two , I got values as X=2 and X=4 , so X hold values from 4 to 2 ? Thanks in advance !
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30 May 2018, 22:37
loserunderachiever wrote: Bunuel wrote: Official Solution:
(1) \(2x^2+9 \lt 9x\). Factor quadratics: \((x\frac{3}{2})(x3) \lt 0\). The roots are \(\frac{3}{2}\) and 3, the "\(\lt\)" sign indicates that the solution lies between the roots: \(1.5 \lt x \lt 3\). Since the only integer in this range is 2, then \(x=2\). Sufficient. (2) \(x+10=2x+8\). The left hand side (LHS) is an absolute value, which is always nonnegative, hence RHS must also be nonnegative: \(2x+8 \ge 0\) giving us \(x \ge 4\). 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.
Answer: D Hi Bunuel , I understood statement A , however , I'm not able to understand statement 2 , in this , X can hold values from 4<x<2 , then X can be 1 also ?? Please guide ,where I'm going wrong. I selected E , while taking the gmat club Quant Test , as I got the same answer as yours , in statement 1, however I missed out X is an integer , in that case statement 1 holds true , in statement two , I got values as X=2 and X=4 , so X hold values from 4 to 2 ? Thanks in advance ! Does plugging x = 1 makes x + 10 = 2x + 8 true? x + 10 = 2x + 8 has only one solution x = 2, I tried to explain (2) in details here: https://gmatclub.com/forum/m27184490.html#p1419035
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