GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 07 Dec 2019, 09:05 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If r and s are the roots of the equation x^2 + bx + c = 0

Author Message
TAGS:

### Hide Tags

Intern  Joined: 11 Sep 2012
Posts: 6
Location: Norway
GMAT Date: 11-19-2013
WE: Brand Management (Pharmaceuticals and Biotech)
If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

### Show Tags

15
137 00:00

Difficulty:   45% (medium)

Question Stats: 64% (01:18) correct 36% (01:42) wrong based on 2529 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

Originally posted by asveaass on 20 Oct 2012, 12:12.
Last edited by Bunuel on 07 Dec 2012, 09:05, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 59588
If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

### Show Tags

34
93
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.

Hope it helps.
_________________
Director  Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 559
Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE: Information Technology (Computer Software)
Re: If r and s are the roots of the equation x^2+bx+c=0, where b  [#permalink]

### Show Tags

22
14
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.

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.

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.
##### General Discussion
VP  Joined: 02 Jul 2012
Posts: 1098
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42 GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: If r and s are the roots of the equation x^2+bx+c=0, where b  [#permalink]

### Show Tags

7
3
Product of roots = c/a = c (in this case)

So question is basically asking if c<0.

Senior Manager  Joined: 15 Jun 2010
Posts: 274
Schools: IE'14, ISB'14, Kellogg'15
WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)
Re: If r and s are the roots of the equation x^2+bx+c=0, where b  [#permalink]

### Show Tags

2
7
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.

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

Director  Joined: 22 Mar 2011
Posts: 584
WE: Science (Education)
Re: If r and s are the roots of the equation x^2+bx+c=0, where b  [#permalink]

### Show Tags

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.

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.
Board of Directors D
Joined: 01 Sep 2010
Posts: 3417
Re: If r and s are the roots of the equation x^2+bx+c=0, where b  [#permalink]

### Show Tags

1
1
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.

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.

Could someone tell me if it is an important, really importante question, or negligible ???
_________________
Board of Directors D
Joined: 01 Sep 2010
Posts: 3417
Re: If r and s are the roots of the equation x^2+bx+c=0, where b  [#permalink]

### Show Tags

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.

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.

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
_________________
Director  Joined: 29 Nov 2012
Posts: 685
Re: If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

### Show Tags

1
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
Manager  Joined: 07 Jan 2010
Posts: 103
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

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
Senior Manager  Joined: 27 Jun 2012
Posts: 347
Concentration: Strategy, Finance
Schools: Haas EWMBA '17
Re: If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

### Show Tags

3
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
VOTE GMAT Practice Tests: Vote Here
PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here
Finance your Student loan through SoFi and get $100 referral bonus : Click here Board of Directors D Joined: 01 Sep 2010 Posts: 3417 Re: What is the value of DS questions [#permalink] ### Show Tags At tthe time this question drove me insane but indeed is fairly stupid read here http://www.purplemath.com/modules/factquad.htm if you need something else ask ;9 _________________ VP  Joined: 02 Jul 2012 Posts: 1098 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Re: What is the value of DS questions [#permalink] ### Show Tags 1 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 = $$\frac{c}{a}$$ Director  S Joined: 09 Jun 2010 Posts: 695 Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink] ### Show Tags gmat dose not require us to remember formular (x-r)(x-s)=x^2+bx +c so, c=rs. B Manager  Joined: 20 Dec 2013 Posts: 222 Location: India Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink] ### Show Tags 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 Intern  Joined: 22 Nov 2013 Posts: 1 Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink] ### Show Tags 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 Manager  Joined: 15 Aug 2013 Posts: 227 Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink] ### Show Tags 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! Math Expert V Joined: 02 Sep 2009 Posts: 59588 Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink] ### Show Tags 2 13 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. _________________ Current Student Joined: 03 Aug 2011 Posts: 277 Concentration: Strategy, Finance GMAT 1: 640 Q44 V34 GMAT 2: 700 Q42 V44 GMAT 3: 680 Q44 V39 GMAT 4: 740 Q49 V41 GPA: 3.7 WE: Project Management (Energy and Utilities) Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink] ### Show Tags 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! Intern  Joined: 31 Jul 2014 Posts: 40 Concentration: Finance, Technology Schools: Owen '17 (M$)
Re: If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

### Show Tags

1
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. Re: If r and s are the roots of the equation x^2 + bx + c = 0   [#permalink] 04 Nov 2014, 12:49

Go to page    1   2    Next  [ 35 posts ]

Display posts from previous: Sort by

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