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# M27-14

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Intern
Joined: 07 Apr 2018
Posts: 6

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24 Jun 2018, 10:13
Bunuel wrote:
p2bhokie wrote:

Bunuel 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 non-negative, 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.

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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. -x-10 = 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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Joined: 02 Sep 2009
Posts: 58427

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24 Jun 2018, 10:29
gmatprep001 wrote:
Bunuel wrote:
p2bhokie wrote:

Bunuel 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 non-negative, 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:
http://gmatclub.com/forum/is-x-157102.html
http://gmatclub.com/forum/what-is-x-126874.html
http://gmatclub.com/forum/is-m-0-1-m-m-2-m-99913.html
http://gmatclub.com/forum/what-is-the-v ... 61134.html
http://gmatclub.com/forum/what-is-the-v ... 69753.html
http://gmatclub.com/forum/what-is-the-v ... 03925.html
http://gmatclub.com/forum/what-is-the-v ... 27976.html
http://gmatclub.com/forum/is-x-0-1-x-3- ... 27978.html
http://gmatclub.com/forum/if-x-0-is-x-1 ... 30402.html
http://gmatclub.com/forum/is-x-0-1-x-3- ... 00357.html
http://gmatclub.com/forum/what-is-the-v ... 45839.html
http://gmatclub.com/forum/the-discreet- ... l#p1039650

Hope this helps.

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. -x-10 = 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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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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26 Sep 2018, 20:05
Bunuel wrote:
Official Solution:

(1) $$2x^2+9 \lt 9x$$. Factor quadratics: $$(x-\frac{3}{2})(x-3) \lt 0$$. 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 non-negative, hence RHS must also be non-negative: $$2x+8 \ge 0$$ 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.

Hi Bunuel
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 non-1 coefficient for x2"
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Joined: 02 Sep 2009
Posts: 58427

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26 Sep 2018, 20:48
1
RamSep wrote:
Bunuel wrote:
Official Solution:

(1) $$2x^2+9 \lt 9x$$. Factor quadratics: $$(x-\frac{3}{2})(x-3) \lt 0$$. 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 non-negative, hence RHS must also be non-negative: $$2x+8 \ge 0$$ 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.

Hi Bunuel
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 non-1 coefficient for x2"

7. Algebra

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Schools: IIMA PGPX"20 (D)
GMAT 1: 640 Q47 V30
GMAT 2: 540 Q42 V22
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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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Joined: 02 Sep 2009
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27 Sep 2018, 23:48
1
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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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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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
-X-10=2X+8
X=-3
Thank you.
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Joined: 02 Sep 2009
Posts: 58427

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09 Oct 2018, 05:25
1
Praveenksinha wrote:
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
-X-10=2X+8
X=-3
Thank you.

Plug back x = -3 does it satisfy the equation?
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09 Oct 2018, 05:29
Bunuel got it, thanks man.
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Joined: 19 Feb 2014
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05 May 2019, 01:14
Bunuel

After factoring as (x-3/2)(x-3)<0, could you please explain why the roots are 3/2 and 3? Isn't the format for factoring (x+factor)(x+factor), and therefore the roots should be -3/2 and -3? Thanks
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05 May 2019, 02:17
crazi4ib wrote:
Bunuel

After factoring as (x-3/2)(x-3)<0, could you please explain why the roots are 3/2 and 3? Isn't the format for factoring (x+factor)(x+factor), and therefore the roots should be -3/2 and -3? Thanks

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05 May 2019, 02:30
Bunuel wrote:
crazi4ib wrote:
Bunuel

After factoring as (x-3/2)(x-3)<0, could you please explain why the roots are 3/2 and 3? Isn't the format for factoring (x+factor)(x+factor), and therefore the roots should be -3/2 and -3? Thanks

Thanks Bunuel , after reading the links I still have the same question. Based on the factored equation, shouldn’t the roots be -3/2 and -3? Thanks

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06 May 2019, 04:48
Re: M27-14   [#permalink] 06 May 2019, 04:48

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# M27-14

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