Another fresh question on 2 Part- Quadratic function : Integrated Reasoning (IR)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 20 Feb 2017, 22:49

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Another fresh question on 2 Part- Quadratic function

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
e-GMAT Representative
Joined: 02 Nov 2011
Posts: 2026
Followers: 2123

Kudos [?]: 7347 [1] , given: 276

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

### Show Tags

25 Jan 2013, 03:56
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
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

Last edited by egmat on 02 Feb 2013, 22:46, edited 1 time in total.
Current Student
Joined: 27 Jun 2012
Posts: 417
Concentration: Strategy, Finance
Followers: 79

Kudos [?]: 786 [4] , given: 184

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

### Show Tags

25 Jan 2013, 16:23
4
KUDOS
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

Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7
VOTE GMAT Practice Tests: Vote Here
PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here

VP
Joined: 08 Jun 2010
Posts: 1416
Followers: 3

Kudos [?]: 119 [0], given: 821

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

### Show Tags

28 Jan 2013, 23:14
how to pick the number ? picking number is time consuming.

any tip, trick here,
e-GMAT Representative
Joined: 02 Nov 2011
Posts: 2026
Followers: 2123

Kudos [?]: 7347 [1] , given: 276

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

### Show Tags

01 Feb 2013, 01:48
1
KUDOS
Expert's post
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
_________________

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Intern
Joined: 19 May 2012
Posts: 36
Location: India
Concentration: International Business, Healthcare
GMAT Date: 03-03-2014
WE: Information Technology (Computer Software)
Followers: 1

Kudos [?]: 7 [0], given: 0

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

### Show Tags

02 Mar 2013, 07:09
x=1,
y=1-root(2)
_________________

Thanks
crazy4priya
GMATPrep 1 710/Q49/V38
GMATPrep 2 690/Q49/V34
Veritas Prep 700/Q50/V36/IR5
MGMT Test 1 700/Q51/V35/IR3

Intern
Joined: 11 May 2013
Posts: 1
Followers: 0

Kudos [?]: 5 [0], given: 0

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

### Show Tags

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
Similar topics Replies Last post
Similar
Topics:
5 A 2 Part question 5 01 Jan 2013, 21:48
4 A tricky 2 Part Question on Multi Level Marketing (MLM) 15 18 Dec 2012, 05:47
3 2-Part IR Question 3 03 Dec 2012, 09:39
2 2 Part question- Max., & Min. value of function! 9 09 Sep 2012, 06:39
2 2 Part question! 2 07 Sep 2012, 06:00
Display posts from previous: Sort by

# Another fresh question on 2 Part- Quadratic function

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.