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Re: M27-14 [#permalink]
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msbcapitalist 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.


Answer: D



Hi Bunuel,

Could you please explain the statement A of the solution. How is the root (x-3/2)?

The roots that I am getting are (x-3) and (2x-3).

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 \((x-3)(x-\frac{3}{2}) \lt 0\).

Theory on Inequalities
Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html
Inequality tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
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Re: M27-14 [#permalink]
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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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: solving-quadratic-inequalities-graphic-approach-170528.html
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Re: M27-14 [#permalink]
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: solving-quadratic-inequalities-graphic-approach-170528.html





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?
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Re: M27-14 [#permalink]
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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: solving-quadratic-inequalities-graphic-approach-170528.html





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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Re: M27-14 [#permalink]
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: the-area-bounded-by-the-curves-x-y-1-and-x-y-1-is-93103.html

can you please explain the difference? where we have take both values and where we have to take only positive value
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Re: M27-14 [#permalink]
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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: the-area-bounded-by-the-curves-x-y-1-and-x-y-1-is-93103.html

can 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.

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html
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Re: M27-14 [#permalink]
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 (x-3) (2x-3)<0 ... I then did the following:

x-3<0 ... so x<3
2x-3<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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Re: M27-14 [#permalink]
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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 (x-3) (2x-3)<0 ... I then did the following:

x-3<0 ... so x<3
2x-3<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: M27-14 [#permalink]
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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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?
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Re: M27-14 [#permalink]
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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.

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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?


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: M27-14 [#permalink]
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.


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 non-1 coefficient for x2"
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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.


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 non-1 coefficient for x2"


Factoring Quadratics
Solving Quadratic Equations


7. Algebra



For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: M27-14 [#permalink]
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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Re: M27-14 [#permalink]
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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: M27-14 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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M27-14 [#permalink]
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.

I also solved statement 2 the conventional way and arrived at X=2 or X=-6. I didn't understand your explanation why this statement would be sufficient until you said:'' you should plug back those values for X, which are 2 or -6, to see if LHD and the RHS of the equation would still be equal''. And yes only when X=2 are the LHS and the RHS equal.

Sometimes are the very tiny details that make the difference in understanding a question.

Thank you Bunuel!
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