It is currently 16 Dec 2017, 01:33

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If r and s are the roots of the equation x^2 + bx + c = 0

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

6 KUDOS received
Intern
Intern
avatar
Joined: 11 Sep 2012
Posts: 6

Kudos [?]: 107 [6], given: 1

Location: Norway
Concentration: Healthcare, International Business
Schools: Insead '14 (A)
GMAT Date: 11-19-2013
WE: Brand Management (Pharmaceuticals and Biotech)
GMAT ToolKit User
If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]

Show Tags

New post 20 Oct 2012, 11:12
6
This post received
KUDOS
58
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

62% (00:57) correct 38% (01:22) wrong based on 2147 sessions

HideShow timer Statistics

If r and s are the roots of the equation x^2 + bx + c = 0, where b and c are constants, is rs < 0 ?

(1) b < 0
(2) c < 0
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 Dec 2012, 08:05, edited 1 time in total.
Renamed the topic and edited the question.

Kudos [?]: 107 [6], given: 1

9 KUDOS received
Director
Director
User avatar
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 635

Kudos [?]: 675 [9], given: 23

Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE: Information Technology (Computer Software)
GMAT ToolKit User Premium Member Reviews Badge
Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]

Show Tags

New post 14 Nov 2012, 09:41
9
This post received
KUDOS
5
This post was
BOOKMARKED
carcass wrote:
Bunuel wrote:
If r and s are the roots of the equation x^2+bx+c=0, where b and c are constants, is rs<0?

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(rs=\frac{c}{1}=\). So, we are basically asked whether \(c<0\).

(1) b<0. Not sufficient.
(2) c<0. Directly answers the question. Sufficient.

Answer: B.

Hope it helps.



This question drove me insane because I didn't really catch the meaning and reasoning. First of all is not mentioned in MGAMT guide, neither in the 5th edition (a copy of the 4th edition) of algebra guide, only in Gmatclub math book a found a reference.

Now from Bunuel I read Viete's theorem.........:(

Indeed also this approach is quite simple

Quote:
If c is positive, then the factors you're looking for are either both positive or else both negative.
If b is positive, then the factors are positive
If b is negative, then the factors are negative.
In either case, you're looking for factors that add to b.

If c is negative, then the factors you're looking for are of alternating signs;
that is, one is negative and one is positive.
If b is positive, then the larger factor is positive.
If b is negative, then the larger factor is negative.
In either case, you're looking for factors that are b units apart.


http://www.purplemath.com/modules/factquad.htm

Could someone tell me if it is an important, really importante question, or negligible ???


Well it would be good to know this concept that in equation ax^2 +bx+c =0
sum of roots =-b/a , product of roots = c/a

Having said that - On GMAT its always test of your reasoning skill. Even if you dont know this you can get the answer.
Lets get back to the question

given is, roots of equation are r and s.
Thus the equation can be written as:
\((x-r)(x-s) =0\)=>
\(x^2 - (r+s)*x +rs =0\)

However question says , equation is x^2+bx+c=0

Comparing the quotients,
b = -(r+s)
C = rs

Now,
statement 1: b <0 => r+s >0 but we cant say anything about rs. Not sufficient.
statement 2: c<0 => rs <0 , exactly what we are looking for. Sufficient

Hence B it is.
_________________

Lets Kudos!!! ;-)
Black Friday Debrief

Kudos [?]: 675 [9], given: 23

Expert Post
8 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42630

Kudos [?]: 135779 [8], given: 12714

Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]

Show Tags

New post 23 Oct 2012, 07:07
8
This post received
KUDOS
Expert's post
44
This post was
BOOKMARKED
If r and s are the roots of the equation x^2+bx+c=0, where b and c are constants, is rs<0?

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(rs=\frac{c}{1}=c\). So, we are basically asked whether \(c<0\).

(1) b<0. Not sufficient.
(2) c<0. Directly answers the question. Sufficient.

Answer: B.

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135779 [8], given: 12714

4 KUDOS received
VP
VP
User avatar
Joined: 02 Jul 2012
Posts: 1214

Kudos [?]: 1720 [4], given: 116

Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Premium Member
Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]

Show Tags

New post 20 Oct 2012, 11:21
4
This post received
KUDOS
3
This post was
BOOKMARKED
Product of roots = c/a = c (in this case)

So question is basically asking if c<0.

Answer is B
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types

Kudos [?]: 1720 [4], given: 116

3 KUDOS received
Current Student
User avatar
Joined: 27 Jun 2012
Posts: 404

Kudos [?]: 962 [3], given: 184

Concentration: Strategy, Finance
Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]

Show Tags

New post 20 Jan 2013, 02:21
3
This post received
KUDOS
gtr022001 wrote:
fozzzy wrote:

In this case you have taken (x - r) (X-s) what would happen if they were both positive (x +r) ( x+s) or even one positive / negative


yes im confused about the oe, the og states: (x-r)(x-s) = x^2-(r+s)x+rs = x^2+bx+c, but shouldn't it be (x+r)(x-s) = x^2+(r+s)x+rs = x^2+bx+c?

in this case (r+s)x = bx?

pls help


fozzzy & gtr022001,

We cannot use \((x+r)(x-s) = x^2+bx+c =0\)
it would mean \((x + r) = 0\) or \((x - s) = 0\), i.e. \(x = -r\) or\(x = s\) --which is inconsistent with information given in the problem

Note that "-r" is not given as a root. Problem states "r" as one of the roots (along with s)
Hence \((x-r)(x-s) = x^2+bx+c =0\)is appropriate with \(x = r\) or \(x = s\) as roots.

Both roots r & s can take any values -> 0, -ve or +ve
if both r & s are either negative or positive, then rs>0
if one of them is positive and other is negative, then rs<0
if both or either one of them is 0, then rs=0

Now refer to Vips0000's explanation that derives \(c=rs\)
as Statement(2) gives \(c<0\), then \(rs<0\)

Hence statement(2) is sufficient to prove \(rs<0\)and answer is choice(B).
_________________

Thanks,
Prashant Ponde

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

Kudos [?]: 962 [3], given: 184

Expert Post
2 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42630

Kudos [?]: 135779 [2], given: 12714

Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]

Show Tags

New post 14 Apr 2014, 01:07
2
This post received
KUDOS
Expert's post
6
This post was
BOOKMARKED
russ9 wrote:
Bunuel wrote:
If r and s are the roots of the equation x^2+bx+c=0, where b and c are constants, is rs<0?

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(rs=\frac{c}{1}=c\). So, we are basically asked whether \(c<0\).

(1) b<0. Not sufficient.
(2) c<0. Directly answers the question. Sufficient.

Answer: B.

Hope it helps.


I actually solved for the discriminant and it ended up taking 3+ minutes. I don't see Viete's theorem in the MGMAT books so please bare with me. I did a search and it basically states that the sum of the two roots is \(\frac{-b}{a}\), and the product of the two roots is \(\frac{c}{a}\), correct?

So using that explanation, it means that since we are looking for the sign of the products of the roots, we need to find the value of \(\frac{c}{a}\) and therefore, \(\frac{(x_1 * x_2)}{a}\). Correct? We aren't given any information about a so how can we assume whether a is positive or negative?

EDIT: Are there similar problems that I can apply this theorem to? Thanks!


\(a\) there is the coefficient of \(x^2\) (\(ax^2+bx+c=0\)), hence for \(x^2+bx+c=0\) it equals to 1 (\(1*x^2+bx+c=0\)).

Questions involving Viete's theorem to practice:
in-the-equation-x-2-bx-12-0-x-is-a-variable-and-b-is-a-109771.html
if-x-2-3-is-one-factor-of-the-equation-x-2-4-3-x-160524.html
if-x-2-12x-k-0-is-x-155465.html
in-the-equation-ax-2-bx-c-0-a-b-and-c-are-constants-148766.html
new-algebra-set-149349-80.html#p1200987
if-q-is-one-root-of-the-equation-x-2-18x-11c-0-where-141199.html
if-f-x-5x-2-and-g-x-x-2-12x-85-what-is-the-sum-of-all-85989.html
if-4-is-one-solution-of-the-equation-x2-3x-k-10-where-139119.html
john-and-jane-started-solving-a-quadratic-equation-john-mad-106597.html

Theory on Algebra: algebra-101576.html

DS Algebra Questions to practice: search.php?search_id=tag&tag_id=29
PS Algebra Questions to practice: search.php?search_id=tag&tag_id=50

Special algebra set: new-algebra-set-149349.html


Hope this helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135779 [2], given: 12714

Expert Post
2 KUDOS received
e-GMAT Representative
User avatar
S
Joined: 04 Jan 2015
Posts: 755

Kudos [?]: 2243 [2], given: 123

Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]

Show Tags

New post 21 Apr 2015, 23:57
2
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
bgpower wrote:
Hi guys,

I couldn't solve the question as I tried to apply the determinant formula of square root (-b +- 4ac), which gave me multiple scenarios. Unfortunately, I didn't know about the Viete's theorem.

Can this question nevertheless be solved based on the above determinant formula?

Thanks!


Hi Guys,

You don't need to know about the Vieta's formula to solve this quadratic equation (if you know, it does save you time :) ). This question can be solved by using the basic application of concepts of quadratic equation.

Let's understand this with an example. Assume a quadratic equation \(x^2 - 8x + 15 =0.\)
We can factorize this as \((x - 5) ( x - 3) = 0\) which will give us the roots as \(x = 5\) and \(x = 3\). We observe here that if 5 and 3 are the roots of a quadratic equation, we can write the equation as \((x -5) ( x-3) = 0\)

Similarly, if r and s are the roots of a quadratic equation, we can write the equation as \((x -r ) ( x - s) = 0\).
So, \((x -r ) ( x - s) = x^2 + bx + c.\)
We can simplify this to \(-x (r +s) + rs = bx + c\). Equation the coefficients we get \(rs = c.\)

So, the question is in effect asking us about the sign of c, which is given by st-II.

Please note that in this case the coefficient of \(x^2\) is 1 which gives us \(rs = c\) and \((r+s) = -b\). Had the coefficient been any other number, lets assume 'a', we would first have to divide the original equation by 'a' to get the coefficient of \(x^2\) as 1.

This would have given us \(rs = \frac{c}{a}\) and \((r+s) = \frac{-b}{a}\) which is what the Vieta's formula tells us :)

Takeaway
Each question can be solved if you know the application of basic concepts the question is checking.

Hope it helps!

Regards
Harsh
_________________












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

Kudos [?]: 2243 [2], given: 123

1 KUDOS received
Board of Directors
User avatar
G
Joined: 01 Sep 2010
Posts: 3424

Kudos [?]: 9523 [1], given: 1203

Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]

Show Tags

New post 13 Nov 2012, 07:34
1
This post received
KUDOS
1
This post was
BOOKMARKED
Bunuel wrote:
If r and s are the roots of the equation x^2+bx+c=0, where b and c are constants, is rs<0?

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(rs=\frac{c}{1}=\). So, we are basically asked whether \(c<0\).

(1) b<0. Not sufficient.
(2) c<0. Directly answers the question. Sufficient.

Answer: B.

Hope it helps.



This question drove me insane because I didn't really catch the meaning and reasoning. First of all is not mentioned in MGAMT guide, neither in the 5th edition (a copy of the 4th edition) of algebra guide, only in Gmatclub math book a found a reference.

Now from Bunuel I read Viete's theorem.........:(

Indeed also this approach is quite simple

Quote:
If c is positive, then the factors you're looking for are either both positive or else both negative.
If b is positive, then the factors are positive
If b is negative, then the factors are negative.
In either case, you're looking for factors that add to b.

If c is negative, then the factors you're looking for are of alternating signs;
that is, one is negative and one is positive.
If b is positive, then the larger factor is positive.
If b is negative, then the larger factor is negative.
In either case, you're looking for factors that are b units apart.


http://www.purplemath.com/modules/factquad.htm

Could someone tell me if it is an important, really importante question, or negligible ???
_________________

COLLECTION OF QUESTIONS AND RESOURCES
Quant: 1. ALL GMATPrep questions Quant/Verbal 2. Bunuel Signature Collection - The Next Generation 3. Bunuel Signature Collection ALL-IN-ONE WITH SOLUTIONS 4. Veritas Prep Blog PDF Version 5. MGMAT Study Hall Thursdays with Ron Quant Videos
Verbal:1. Verbal question bank and directories by Carcass 2. MGMAT Study Hall Thursdays with Ron Verbal Videos 3. Critical Reasoning_Oldy but goldy question banks 4. Sentence Correction_Oldy but goldy question banks 5. Reading-comprehension_Oldy but goldy question banks

Kudos [?]: 9523 [1], given: 1203

1 KUDOS received
Current Student
avatar
Joined: 31 Jul 2014
Posts: 41

Kudos [?]: 20 [1], given: 0

Concentration: Finance, Technology
Schools: Owen '17 (M)
GMAT ToolKit User
Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]

Show Tags

New post 04 Nov 2014, 11:49
1
This post received
KUDOS
Do not need any formula to solve...this question is actually relatively easy. Just think about what it means for r and s to be roots in a quadratic.

1.) b < 0

In order for this to be true we can have the case of (x+r)(x-s) which would result in rs being negative. Or we can have the case of (x-r)(x-s) which would result in rs being positive. Therefore this cannot be sufficient.

2.) s<0

If s is negative it means either r is negative by the case (x-r)(x+s) or s is negative by the case (x+r)(x-s). In either case the result is rs is negative. Therefore this is sufficient.

Kudos [?]: 20 [1], given: 0

Expert Post
1 KUDOS received
Math Expert
User avatar
D
Joined: 02 Aug 2009
Posts: 5354

Kudos [?]: 6141 [1], given: 121

Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]

Show Tags

New post 23 May 2016, 05:56
1
This post received
KUDOS
Expert's post
nishi999 wrote:
I don't understand why have you considered the roots (x-r) and (x-s). Couldn't it even be (x + r) & (x + s)


Hi,
if roots are r and s, then \((x-r)(x-s)=0\)is correct...
If roots were -r and -s, then \((x+r)(x+s)=0..\)

Reason is at the value of roots, the equation becomes 0...
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6141 [1], given: 121

Senior Manager
Senior Manager
avatar
Joined: 15 Jun 2010
Posts: 356

Kudos [?]: 468 [0], given: 50

Schools: IE'14, ISB'14, Kellogg'15
WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)
Reviews Badge
Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]

Show Tags

New post 21 Oct 2012, 01:08
6
This post was
BOOKMARKED
asveaass wrote:
Thanks, but could elaborate on your steps towards your answer please? :)

MacFauz wrote:
Product of roots = c/a = c (in this case)

So question is basically asking if c<0.

Answer is B


We know
1) Sum of roots(r,s) of a Quad Equation in the form ax^2+bx+c can be written as=> r+s = -(b/a)
2) Product of roots => rs = c/a

Now in this case a=1, So Product of roots rs = c. So the question asks is rs<0 ? i.e, by rephrasing we get is c<0??

St 2: Sufficient

Hence Answer B
_________________

Regards
SD
-----------------------------
Press Kudos if you like my post.
Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html

Kudos [?]: 468 [0], given: 50

Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 610

Kudos [?]: 1091 [0], given: 43

WE: Science (Education)
Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]

Show Tags

New post 21 Oct 2012, 04:32
asveaass wrote:
Thanks, but could elaborate on your steps towards your answer please? :)

MacFauz wrote:
Product of roots = c/a = c (in this case)

So question is basically asking if c<0.

Answer is B


Relevant post:

if-the-graph-of-y-x-2-ax-b-passes-through-the-points-139828.html#p1131650
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Kudos [?]: 1091 [0], given: 43

Board of Directors
User avatar
G
Joined: 01 Sep 2010
Posts: 3424

Kudos [?]: 9523 [0], given: 1203

Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]

Show Tags

New post 14 Nov 2012, 10:02
Vips0000 wrote:
carcass wrote:
Bunuel wrote:
If r and s are the roots of the equation x^2+bx+c=0, where b and c are constants, is rs<0?

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(rs=\frac{c}{1}=\). So, we are basically asked whether \(c<0\).

(1) b<0. Not sufficient.
(2) c<0. Directly answers the question. Sufficient.

Answer: B.

Hope it helps.


It's a shortcut. I also agree with you to spot the sense of the meaning of the question only thinking that rs <0 have opposite sign :)

Good to know eventhough what i said is even faster.

Thanks :)


This question drove me insane because I didn't really catch the meaning and reasoning. First of all is not mentioned in MGAMT guide, neither in the 5th edition (a copy of the 4th edition) of algebra guide, only in Gmatclub math book a found a reference.

Now from Bunuel I read Viete's theorem.........:(

Indeed also this approach is quite simple

Quote:
If c is positive, then the factors you're looking for are either both positive or else both negative.
If b is positive, then the factors are positive
If b is negative, then the factors are negative.
In either case, you're looking for factors that add to b.

If c is negative, then the factors you're looking for are of alternating signs;
that is, one is negative and one is positive.
If b is positive, then the larger factor is positive.
If b is negative, then the larger factor is negative.
In either case, you're looking for factors that are b units apart.


http://www.purplemath.com/modules/factquad.htm

Could someone tell me if it is an important, really importante question, or negligible ???


Well it would be good to know this concept that in equation ax^2 +bx+c =0
sum of roots =-b/a , product of roots = c/a

Having said that - On GMAT its always test of your reasoning skill. Even if you dont know this you can get the answer.
Lets get back to the question

given is, roots of equation are r and s.
Thus the equation can be written as:
\((x-r)(x-s) =0\)=>
\(x^2 - (r+s)*x +rs =0\)

However question says , equation is x^2+bx+c=0

Comparing the quotients,
b = -(r+s)
C = rs

Now,
statement 1: b <0 => r+s >0 but we cant say anything about rs. Not sufficient.
statement 2: c<0 => rs <0 , exactly what we are looking for. Sufficient

Hence B it is.


Of course I know how to get the roots or the use of the discriminant, but in different source that I have this odd angle of the question is not mentioned. Good to know

Got it :)

Thanks
_________________

COLLECTION OF QUESTIONS AND RESOURCES
Quant: 1. ALL GMATPrep questions Quant/Verbal 2. Bunuel Signature Collection - The Next Generation 3. Bunuel Signature Collection ALL-IN-ONE WITH SOLUTIONS 4. Veritas Prep Blog PDF Version 5. MGMAT Study Hall Thursdays with Ron Quant Videos
Verbal:1. Verbal question bank and directories by Carcass 2. MGMAT Study Hall Thursdays with Ron Verbal Videos 3. Critical Reasoning_Oldy but goldy question banks 4. Sentence Correction_Oldy but goldy question banks 5. Reading-comprehension_Oldy but goldy question banks

Kudos [?]: 9523 [0], given: 1203

Director
Director
avatar
Joined: 29 Nov 2012
Posts: 862

Kudos [?]: 1492 [0], given: 543

Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]

Show Tags

New post 14 Jan 2013, 20:02
1
This post was
BOOKMARKED
Vips0000 wrote:

Well it would be good to know this concept that in equation ax^2 +bx+c =0
sum of roots =-b/a , product of roots = c/a

Having said that - On GMAT its always test of your reasoning skill. Even if you dont know this you can get the answer.
Lets get back to the question

given is, roots of equation are r and s.
Thus the equation can be written as:
\((x-r)(x-s) =0\)=>
\(x^2 - (r+s)*x +rs =0\)

However question says , equation is x^2+bx+c=0

Comparing the quotients,
b = -(r+s)
C = rs

Now,
statement 1: b <0 => r+s >0 but we cant say anything about rs. Not sufficient.
statement 2: c<0 => rs <0 , exactly what we are looking for. Sufficient

Hence B it is.


In this case you have taken (x - r) (X-s) what would happen if they were both positive (x +r) ( x+s) or even one positive / negative
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Kudos [?]: 1492 [0], given: 543

Manager
Manager
User avatar
Joined: 07 Jan 2010
Posts: 143

Kudos [?]: 97 [0], given: 57

Location: So. CA
WE 1: 2 IT
WE 2: 4 Software Analyst
Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]

Show Tags

New post 19 Jan 2013, 13:59
fozzzy wrote:

In this case you have taken (x - r) (X-s) what would happen if they were both positive (x +r) ( x+s) or even one positive / negative


yes im confused about the oe, the og states: (x-r)(x-s) = x^2-(r+s)x+rs = x^2+bx+c, but shouldn't it be (x+r)(x-s) = x^2+(r+s)x+rs = x^2+bx+c?

in this case (r+s)x = bx?

pls help

Kudos [?]: 97 [0], given: 57

Board of Directors
User avatar
G
Joined: 01 Sep 2010
Posts: 3424

Kudos [?]: 9523 [0], given: 1203

Re: What is the value of DS questions [#permalink]

Show Tags

New post 14 Feb 2013, 17:05

Kudos [?]: 9523 [0], given: 1203

VP
VP
User avatar
Joined: 02 Jul 2012
Posts: 1214

Kudos [?]: 1720 [0], given: 116

Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Premium Member
Re: What is the value of DS questions [#permalink]

Show Tags

New post 14 Feb 2013, 20:47
Val1986 wrote:
Hey Guys,

Stubled upon the following question:

If r and s are the roots of the equation x^2 + bx + c = 0, where b and c are constancts, is rs < 0?

1) b<0

2) c <0

Any takers?


For any equation, \(ax^2 + bx + c = 0\)

Sum of roots = \(-\frac{b}{a}\)

Product of roots = \(\frac{c}{a}\)
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types

Kudos [?]: 1720 [0], given: 116

VP
VP
avatar
S
Joined: 08 Jun 2010
Posts: 1387

Kudos [?]: 172 [0], given: 916

Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]

Show Tags

New post 19 Feb 2013, 08:15
gmat dose not require us to remember formular

(x-r)(x-s)=x^2+bx +c

so, c=rs.

B
_________________

visit my facebook to help me.
on facebook, my name is: thang thang thang

Kudos [?]: 172 [0], given: 916

Senior Manager
Senior Manager
avatar
Joined: 20 Dec 2013
Posts: 267

Kudos [?]: 111 [0], given: 29

Location: India
Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]

Show Tags

New post 04 Mar 2014, 02:53
rs=product of roots.
In this case product of roots=c
In S2 we're directly given c<0 which implies rs<0.Hence S2 is sufficient.

S1:b<0 doesn't provide useful info.

Posted from my mobile device

Kudos [?]: 111 [0], given: 29

Intern
Intern
avatar
Joined: 22 Nov 2013
Posts: 1

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

Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]

Show Tags

New post 04 Mar 2014, 10:56
With the traditional formula to solve quadratic equation:

(b-\sqrt{b^2-4c})/2 ("a" omitted, since a=1)

The question asks if "r" or "s" is negative.

One of the solutions is negative only when \sqrt{b^2-4c} > b that is

b^2-4c>b^2

-4c>0

c>0

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

Re: If r and s are the roots of the equation x^2 + bx + c = 0   [#permalink] 04 Mar 2014, 10:56

Go to page    1   2    Next  [ 32 posts ] 

Display posts from previous: Sort by

If r and s are the roots of the equation x^2 + bx + c = 0

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


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

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®.