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Re: M2714
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24 Jun 2018, 10:13
Bunuel wrote: Thanks Bunuel. Could you quickly confirm my understanding: 1) Test positive case for x+10, i.e. x+10 = 2x+8, i.e. x = 2 2) Test negative case for x+10, i.e. x10 = 2x+8, i.e. x = 6 => not a solution because of constraint x >= 4 Thus, x = 2 as only solution, right?



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24 Jun 2018, 10:29
gmatprep001 wrote: Bunuel wrote: Thanks Bunuel. Could you quickly confirm my understanding: 1) Test positive case for x+10, i.e. x+10 = 2x+8, i.e. x = 2 2) Test negative case for x+10, i.e. x10 = 2x+8, i.e. x = 6 => not a solution because of constraint x >= 4 Thus, x = 2 as only solution, right? If you figured that x must be more than or equal to 4, then you could directly get that x + 10 = x + 10 because if x >= 4, then x + 10 is positive, thus x + 10 = x + 10 (recall that a = a if a >= 0). If you want to open modulus in conventional way then you should consider two cases: x <= 10 and x >= 10.
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Re: M2714
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26 Sep 2018, 07:19
Bunuel wrote: (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 for the explanation. This is exactly where I had got myself wrong. (my solution was x<3/2 and X<3)
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Re: M2714
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26 Sep 2018, 20:05
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 Thanks for your explanations I have a real problem with factorizing quadratics containing "a>1 or a<1" coefficient (ax2+bx+c). To me it's a real headache & nightmare, making me sometimes to use formula to find the roots which is absolutely time consuming and under the pressure of the test, mostly doesn't work at all. I would be grateful if you share me anything on "how to factorize a quadratic equation with a non1 coefficient for x2"



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26 Sep 2018, 20:48
RamSep 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 Thanks for your explanations I have a real problem with factorizing quadratics containing "a>1 or a<1" coefficient (ax2+bx+c). To me it's a real headache & nightmare, making me sometimes to use formula to find the roots which is absolutely time consuming and under the pressure of the test, mostly doesn't work at all. I would be grateful if you share me anything on "how to factorize a quadratic equation with a non1 coefficient for x2" Factoring QuadraticsSolving Quadratic Equations 7. Algebra For more check Ultimate GMAT Quantitative MegathreadHope it helps.
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Re: M2714
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27 Sep 2018, 23:33
Bunuel For Statement B : x+10=2x+8 ,we can consider x+10= 2x+8 and x+10 =  [2x+8] and this gives 2 values 2 and 6 and so we can say B is not sufficient. I also understand your and other justification how B can be correct as well but why I am posting question is because I have seen other problems and solution on our club where when given mod of X , we consider both values of X while doing calculations. So want to understand when to consider both values + and  and when to consider only + value ?



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27 Sep 2018, 23:48
NAvinash wrote: Bunuel For Statement B : x+10=2x+8 ,we can consider x+10= 2x+8 and x+10 =  [2x+8] and this gives 2 values 2 and 6 and so we can say B is not sufficient. I also understand your and other justification how B can be correct as well but why I am posting question is because I have seen other problems and solution on our club where when given mod of X , we consider both values of X while doing calculations. So want to understand when to consider both values + and  and when to consider only + value ? When you solve the way you did, you should plug back the roots to confirm that they do in fact satisfy the equation. If you plug x = 6 you'd get that LHS = 4 and RHS = 4. 4 ≠ 4. So, x = 6 does not satisfy x + 10 = 2x + 8. Or if you adopt more conventional way you'd have: If x < 10, then x + 10 < 0, so x + 10 = (x + 10) and in this case we'd have (x + 10) = 2x + 8 > x = 6. You should discard this solution because x = 6 is not in the range we are considering (x < 10). If x >= 10, then x + 10 >= 0, so x + 10 = x + 10 and in this case we'd have x + 10 = 2x + 8 > x = 2. This solution is valid because x = 2 is in the range we are considering (x >= 10). So, x + 10 = 2x + 8 has only one solution: x = 2. Hope it's clear.
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Re: M2714
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02 Oct 2018, 20:28
Thanks for the last post/reply Bunuel. This helps. I will prefer first part of answer as it is easy to apply in exam. I will solve LHS RHS using the roots obtained as one of the easiest and quick trial way. Thanks.



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Re: M2714
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09 Oct 2018, 05:17
Hello Bunuel I have doubt in second statement need your help. I have solved it by taking two condition 1) when X>=10 in this case equation became x+10=2x+8 X=2 case 2) when X<10 X10=2X+8 X=3 hence Two solution not sufficient, please help me to get this right. Thank you.



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Re: M2714
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09 Oct 2018, 05:25



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Re: M2714
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09 Oct 2018, 05:29
Bunuel got it, thanks man.







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