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Very Tough Algebra II Question  Please Help [#permalink]
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14 Oct 2009, 23:17
If A and B are integers and B = A + 4, which of the following represents integer X for which the expression (X  A)^2 + (X  B)^2 is the smallest? A) A1 B) A C) A+2 D) A+3 E) B+1 OA Can anyone help me understand this question?



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Re: Very Tough Algebra II Question  Please Help [#permalink]
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14 Oct 2009, 23:37
dpgxxx wrote: If A and B are integers and B = A + 4, which of the following represents integer X for which the expression (X  A)^2 + (X  B)^2 is the smallest? A) A1 B) A C) A+2 D) A+3 E) B+1 OA Can anyone help me understand this question? One can make out that for expression (X  A)^2 + (X  B)^2 to be smallest we need to select value for X such that the squares are having least value For X=A we get (AA)^2 + (AA4)^2 = 16 For X = A+2 we get (A+2 A)^2 + (A+2 A4)^2 = 8 For X= A+3 we get (A+3 A)^2 + (A+3 A4)^2 = 10 for expression A1 and B+1 we will get value more than 8 so I will go with X=A+2. Option C as answer



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Re: Very Tough Algebra II Question  Please Help [#permalink]
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19 Oct 2009, 21:04
I go along with the explanation above, which is the easiest way to SOLVE the problem but not to UNDERSTAND it.
My understanding is like this:
Since B=X+4, the expression (XA)^2+(XB)^2 =(XA)^2+[(XA)4)]^2 =(XA)^2+(XA)^28(XA)+16 =2[(XA)^24(XA)+8] =2[(XA)^24(XA)+4+4] =2[(XA2)^2+4]
As (XA2)^2 can not be less than 0, the expression is smallest when (XA2)^2 =0, Therefore, option C: X=A+2 is selected
The reasoning process is only for understanding the problem, I still recommend the method by asterixmatrix for problemsolving purpose in GMAT testing



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Re: Very Tough Algebra II Question  Please Help [#permalink]
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19 Oct 2009, 22:57
dpgxxx wrote: If A and B are integers and B = A + 4, which of the following represents integer X for which the expression (X  A)^2 + (X  B)^2 is the smallest? A) A1 B) A C) A+2 D) A+3 E) B+1 OA Can anyone help me understand this question? Given that: (X  A)^2 + (X  B)^2 = (X  A)^2 + (X  a  4)^2 If x = a1, (X  A)^2 part becomes 1 but (X  a  4)^2 part becomes 25. So total = 26. If x = a, (X  A)^2 part becomes 0 but (X  a  4)^2 part becomes 16. So total = 16. If x = a+2, (X  A)^2 part becomes 4 but (X  a  4)^2 part becomes 4. So total = 8.If x = a+3, (X  A)^2 part becomes 9 but (X  a  4)^2 part becomes 1. So total = 10. If x = b+1=a+4+1=a+5, (X  A)^2 part becomes 25 and (X  a  4)^2 = 1 So total = 26 Thats C.
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Re: Very Tough Algebra II Question  Please Help [#permalink]
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20 Oct 2009, 12:56
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dpgxxx wrote: If A and B are integers and B = A + 4, which of the following represents integer X for which the expression (X  A)^2 + (X  B)^2 is the smallest? A) A1 B) A C) A+2 D) A+3 E) B+1 OA Can anyone help me understand this question? 18 seconds method, take the derivatives of both polynomials of order two to get 2 ( 2x  A  B) = 0, now substitute for B as given in the question and get the value of X which is X = A + 2



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Re: Very Tough Algebra II Question  Please Help [#permalink]
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20 Oct 2009, 21:52
LUGO wrote: 18 seconds method, take the derivatives of both polynomials of order two to get 2 ( 2x  A  B) = 0, now substitute for B as given in the question and get the value of X which is X = A + 2 Lugo, thanks for reminding me about derivatives! Such a basic concept, completely overlooked on gmat prepping. To find the minimum/maximum value of an equation, find the point where its derivative is 0. Given that this is a convex (upwards function), the value obtained will be the minimum value. Some useful links for going over derivatives: http://www.1728.com/minmax.htmhttp://www.tutorvista.com/content/math/ ... minima.phphttp://en.wikipedia.org/wiki/DerivativeCheers
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Re: Very Tough Algebra II Question  Please Help [#permalink]
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06 Oct 2010, 13:45
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I solved it using derivative. But do not forget to check the second derivative. If second derivate is >0 then only the first derivate can be simplified to get the min value. Another solution: \((xa)^2 + (xb)^2 =\) square of the distance between (x,x) and (a,b) Since (x,x) lies on the line x=y the min distance between the line x=y and (a,b) is our required result. This distance will be minimum when the perpendicular from the point a,b intersects the x=y at point x,x since slope of x=y is 1, the slope of perpendicular from a,b to line x=y is 1 \(\frac{(xa)}{(xb)} = 1\) \(xa = bx\) \(x = \frac{(a+b)}{2}\) \(x = \frac{(a+a+4)}{2} = a+2\) Hence C
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Re: M19 Q18 [#permalink]
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07 Sep 2011, 13:46
krishnasty wrote: If A and B are integers and B=A+4 , which of the following represents integer x for which the expression (xA)^2 + (xB)^2 is the smallest?
1. A1 2. A 3. A+2 4. A+3 5. B+1 Nice question and it can be answered in 10 sec  Answer is 3. A+2 for the expression to be smallest, Integer x must reside between B and A, hence (B+A)/2 = A+2. However, plug and play can also be applied to resolve this question. Hope it helps. Aj.
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Re: M19 Q18 [#permalink]
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13 Sep 2011, 01:38
how you know this "for the expression to be smallest, Integer x must reside between B and A", could u plz explain.. is it bcuz this especial expression? thnx



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Re: M19 Q18 [#permalink]
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13 Sep 2011, 06:12
krishnasty wrote: If A and B are integers and B=A+4 , which of the following represents integer x for which the expression (xA)^2 + (xB)^2 is the smallest?
1. A1 2. A 3. A+2 4. A+3 5. B+1 since (xA)^2 + (xB)^2 and B=A+4 we have (xA)^2 + (x(A+4))^2 or (xA)^2 + (xA4)^2 plug in each answ.choice and u get the answ i.e. x=a1 then ((A1)A)^2 + ((A1)A4)^2 =1+25=26 and so on ...each answ choise...
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Re: M19 Q18 [#permalink]
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13 Sep 2011, 07:22
mehrasa wrote: how you know this "for the expression to be smallest, Integer x must reside between B and A", could u plz explain.. is it bcuz this especial expression? thnx Yes it is only applicable to this special expression. If Integer x doesn't reside between B and A than one of the value (either (xA)^2 or (xB)^2) in expression (xA)^2 + (xB)^2 will explode. Hope it helps. Cheers!
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Re: Very Tough Algebra II Question  Please Help [#permalink]
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18 Mar 2012, 02:39
gurpreetsingh wrote: This distance will be minimum when the perpendicular from the point a,b intersects the x=y at point x,x
I am not able to understand why it is so?



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Re: M19 Q18 [#permalink]
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26 Apr 2013, 23:12
(I've posted another approach to this problem below this post) Hi My friends, (sorry for my bad English) This problem will be much easier and only take 10 seconds to be solved if you know this inequality : a^2+b^2>= 2 a.b (proof: we have (ab)^2 >=0 so please expand it and then we have a^2 +b^2 >= 2 a.b). ( this is AMGM inequalityvery popular and basic inequality) So the minimum of a^2 +b^2 will be 2 a.b only when a=b. .....and applied to this problem, we will have the minimum value when xA=xB so we have x= (A+B)/2= A+2 The answer is C
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Re: Very Tough Algebra II Question  Please Help [#permalink]
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27 Apr 2013, 04:38
dpgxxx wrote: If A and B are integers and B = A + 4, which of the following represents integer X for which the expression (X  A)^2 + (X  B)^2 is the smallest? A) A1 B) A C) A+2 D) A+3 E) B+1 OA Can anyone help me understand this question? If \(b=a+4\), then for which of the following values of \(x\) is the expression \((xa)^2 + (xb)^2\) the smallest?A. \(a1\) B. \(a\) C. \(a+2\) D. \(a+3\) E. \(a+5\) Since \(b=a+4\) then \((xa)^2 + (xb)^2=(xa)^2 + (xa4)^2\). Now, plug each option for \(x\) to see which gives the least value. The least value of the expression if for \(x=a+2\) > \((xa)^2 + (xa4)^2=(a+2a)^2 + (a+2a4)^2=8\). Answer: C.
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Re: M19 Q18 [#permalink]
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27 Apr 2013, 12:20
Hi my friends, let me try another approach to this problem: let's \(C=(xA)^2 + (xB)^2\) so \(C= 2x^2 +A^2+B^2  2 x. (A+B)\) and then \(C = (\sqrt{2}x)^2 + \frac{(A+B)^2}{2} + \frac{(AB)^2}{2}  2.\sqrt{2}.x. \frac{(A+B)}{\sqrt{2}}\) Why I do this? cause I want to simplified this to \(x^2+P >= P\) now we have \(C= (\sqrt{2}x\frac{(A+B)}{\sqrt{2}})^2 + \frac{(AB)^2}{2}\) clearly \(C>= \frac{(AB)^2}{2}=8\) \(C=8\) (minimum) only when \(\sqrt{2}.x=\frac{(A+B)}{\sqrt{2}}\) and then we have \(x=\frac{(A+B)}{2} = A+2\) The answer: C
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