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Another fresh question on 2 Part- Quadratic function

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Another fresh question on 2 Part- Quadratic function [#permalink] New post 25 Jan 2013, 03:56
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Try yet another 2 part question- fresh from the e-GMAT bakery!

A function f(x, y) is such that f(x,y)=3x^2-2xy+y^2+4. Select one value for x, & one value for y such that given information implies that f(x, y) = 8. Make only two selections, one in each column.

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-Shalabh

Last edited by egmat on 02 Feb 2013, 22:46, edited 1 time in total.
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Re: Another fresh question on 2 Part- Quadratic function [#permalink] New post 25 Jan 2013, 16:23
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Answer: x = 1 and y= 1-\sqrt{2}

I used back-solving method to solve this problem by substituting values for x in following order (0, 1, -1, 1-\sqrt{2})

Given that,
f(x,y)=3x^2-2xy+y^2+4=8
i.e. 3x^2-2xy+y^2=4 -- To be proved

Substituting x=0 gives y=\pm2 which is not in the answer list.

Substitute x=1
3x^2-2xy+y^2=4
3*1^2-2(1)y+y^2=4
3-2y+y^2=4
y^2-2y-1=0

As we know x = [-b\pm\sqrt{b^2-4ac}]/2a are roots for ax^2+bx+c=0

y= [-(-2)\pm\sqrt{(-2)^2-4(1)(-1)}]/2(1)=[2\pm\sqrt{(8)}]/2=1-\sqrt{2}


Hence Answer: x = 1 and y= 1-\sqrt{2}
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Re: Another fresh question on 2 Part- Quadratic function [#permalink] New post 28 Jan 2013, 23:14
how to pick the number ? picking number is time consuming.

any tip, trick here,
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Re: Another fresh question on 2 Part- Quadratic function [#permalink] New post 01 Feb 2013, 01:48
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thangvietnam wrote:
how to pick the number ? picking number is time consuming.

any tip, trick here,

Hi,

When we look at f(x,y)=3x^2-2xy+y^2+4 and then at the options, we find that plugging values is the best way to approach this question.
There are 3 things that should be keptp in mind while picking option values.

1.Pick integers first. They are easy to work on.
There are 3 values in the option list, which are integers.

2.Pick ‘0’ first. This will eliminate one variable completely for compuation.

3.What to choose first; x or y? One should always observe right hand side of the function. If number of terms of x is more than the number of terms of y, then plug in the option value in x first, and vice versa.


We choose the values for x in the order of 0, 1, and -1 to plug in.

Now, we plug in the value of f(x, y) =8, & x=0 in the equation, and we get,

8=3.0^2-2.0.y+y^2+4
8=y^2+4
4=y^2
y= ±2

This means for x=0, y is either 2 or -2. There is no such option available for values: 2 or-2 , hence these pair of values cannot be correct.

Now, we should try x=1. You may follow PraPon’s solution for x=1. He has done it correctly.

Hope it helps!

-Shalabh
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Re: Another fresh question on 2 Part- Quadratic function [#permalink] New post 02 Mar 2013, 07:09
x=1,
y=1-root(2)
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Re: Another fresh question on 2 Part- Quadratic function [#permalink] New post 10 Nov 2013, 12:15
f(x,y) = 3x^2 - 2xy + y^2 + 4 = 2x^2 + (x-y)^2 + 4
So, f(x,y) = 8 if and only if 2x^2 + (x-y)^2 = 4 or x^2 + ((x-y)/sqrt 2)^2 = 2

Now that you have formulated the expression this way, it is very easy to see that x=1 and y=1-sqrt(2) is the solution.

This is much faster than plugging in values.
Re: Another fresh question on 2 Part- Quadratic function   [#permalink] 10 Nov 2013, 12:15
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Another fresh question on 2 Part- Quadratic function

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