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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 (A-A)^2 + (A-A-4)^2 = 16 For X = A+2 we get (A+2 -A)^2 + (A+2 -A-4)^2 = 8 For X= A+3 we get (A+3 -A)^2 + (A+3 -A-4)^2 = 10 for expression A-1 and B+1 we will get value more than 8

Re: Very Tough Algebra II Question --- Please Help [#permalink]

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19 Oct 2009, 20:04

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 (X-A)^2+(X-B)^2 =(X-A)^2+[(X-A)-4)]^2 =(X-A)^2+(X-A)^2-8(X-A)+16 =2[(X-A)^2-4(X-A)+8] =2[(X-A)^2-4(X-A)+4+4] =2[(X-A-2)^2+4]

As (X-A-2)^2 can not be less than 0, the expression is smallest when (X-A-2)^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 problem-solving purpose in GMAT testing

Given that: (X - A)^2 + (X - B)^2 = (X - A)^2 + (X - a - 4)^2

If x = a-1, (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

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

Re: Very Tough Algebra II Question --- Please Help [#permalink]

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20 Oct 2009, 20:52

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.

Re: Very Tough Algebra II Question --- Please Help [#permalink]

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06 Oct 2010, 12:45

2

This post received KUDOS

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:

\((x-a)^2 + (x-b)^2 =\) 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

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?

1. A-1 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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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

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?

1. A-1 2. A 3. A+2 4. A+3 5. B+1

since (x-A)^2 + (x-B)^2 and B=A+4 we have (x-A)^2 + (x-(A+4))^2 or (x-A)^2 + (x-A-4)^2 plug in each answ.choice and u get the answ i.e. x=a-1 then ((A-1)-A)^2 + ((A-1)-A-4)^2 =1+25=26 and so on ...each answ choise...
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I am still on all gmat forums. msg me if you want to ask me smth

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 (x-A)^2 or (x-B)^2) in expression (x-A)^2 + (x-B)^2 will explode.

Hope it helps. Cheers!
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(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 (a-b)^2 >=0 so please expand it and then we have a^2 +b^2 >= 2 a.b). ( this is AM-GM inequality-very 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 x-A=x-B so we have x= (A+B)/2= A+2 The answer is C
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Last edited by retailingvnsupernova on 27 Apr 2013, 11:21, edited 1 time in total.

Hi my friends, let me try another approach to this problem: let's \(C=(x-A)^2 + (x-B)^2\) so \(C= 2x^2 +A^2+B^2 - 2 x. (A+B)\) and then \(C = (\sqrt{2}x)^2 + \frac{(A+B)^2}{2} + \frac{(A-B)^2}{2} - 2.\sqrt{2}.x. \frac{(A+B)}{\sqrt{2}}\) Why I do this? cause I want to simplified this to \(x^2+P >= P\) now we have \(C= (\sqrt{2}x-\frac{(A+B)}{\sqrt{2}})^2 + \frac{(A-B)^2}{2}\) clearly \(C>= \frac{(A-B)^2}{2}=8\) \(C=8\) (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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