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# Another fresh question on 2 Part- Quadratic function

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e-GMAT Representative
Joined: 02 Nov 2011
Posts: 2711
Another fresh question on 2 Part- Quadratic function  [#permalink]

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Updated on: 02 Feb 2013, 23:46
1
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.

This is a representative question of OG13/# 38. Want to view similar 2 part questions with an interactive audio visual solution? Register here at e-GMAT.

-Shalabh

Originally posted by egmat on 25 Jan 2013, 04:56.
Last edited by egmat on 02 Feb 2013, 23:46, edited 1 time in total.
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Re: Another fresh question on 2 Part- Quadratic function  [#permalink]

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25 Jan 2013, 17:23
5
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 $$y= 1-\sqrt{2}$$
_________________

Thanks,
Prashant Ponde

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Re: Another fresh question on 2 Part- Quadratic function  [#permalink]

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29 Jan 2013, 00:14
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]

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01 Feb 2013, 02:48
2
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]

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02 Mar 2013, 08:09
x=1,
y=1-root(2)
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Thanks
crazy4priya
GMATPrep 1 710/Q49/V38
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Re: Another fresh question on 2 Part- Quadratic function  [#permalink]

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10 Nov 2013, 13:15
$$f(x,y) = 3x^2 - 2xy + y^2 + 4 = 2x^2 + (x-y)^2 + 4$$
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.
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Re: Another fresh question on 2 Part- Quadratic function  [#permalink]

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25 Nov 2017, 06:32
Here is another trick as number picking can be a longer process !
Solve and separate out for quadratic
we know we have to prove 3x^2-2xy+y^2=4
i will simplify this as 2x^2+(x-y)^2=4 ...... from here choosing x =1 and y = 1-root2 becomes easy

Kudos if it helps
Re: Another fresh question on 2 Part- Quadratic function &nbs [#permalink] 25 Nov 2017, 06:32
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# Another fresh question on 2 Part- Quadratic function

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