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If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]
 20 Oct 2012, 12:12
Question Stats:
50% (01:33) correct
50% (01:02) wrong based on 6 sessions
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
Last edited by Bunuel on 07 Dec 2012, 09:05, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]
20 Oct 2012, 12:21
Product of roots = c/a = c (in this case) So question is basically asking if c<0. Answer is B
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]
21 Oct 2012, 01:54
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
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]
21 Oct 2012, 02:08
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
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]
21 Oct 2012, 05:32
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
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]
23 Oct 2012, 08:07
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]
13 Nov 2012, 08:34
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.htmCould someone tell me if it is an important, really importante question, or negligible ???
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]
14 Nov 2012, 10:41
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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.htmCould 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 =0However 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.
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink]
14 Nov 2012, 11:02
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.htmCould 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 =0However 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
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Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]
14 Jan 2013, 21:02
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
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Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]
19 Jan 2013, 14:59
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
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Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]
20 Jan 2013, 03:21
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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
We cannot use (x+r)(x-s) = x^2+bx+c =0it would mean (x + r) = 0 or (x - s) = 0, i.e. x = -r or x = s --which is inconsistent with information given in the problemNote 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 =0is 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=rsas Statement(2) gives c<0, then rs<0Hence statement(2) is sufficient to prove rs<0and answer is choice(B).
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What is the value of DS questions [#permalink]
14 Feb 2013, 17:55
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?
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Re: What is the value of DS questions [#permalink]
14 Feb 2013, 18:05
At tthe time this question drove me insane but indeed is fairly stupid read here http://www.purplemath.com/modules/factquad.htmif you need something else ask ;9
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Re: What is the value of DS questions [#permalink]
14 Feb 2013, 21:47
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 = 0Sum of roots = -\frac{b}{a}Product of roots = \frac{c}{a}
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Re: What is the value of DS questions [#permalink]
15 Feb 2013, 02:13
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Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink]
19 Feb 2013, 09:15
gmat dose not require us to remember formular
(x-r)(x-s)=x^2+bx +c
so, c=rs.
B
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Re: If r and s are the roots of the equation x^2 + bx + c = 0
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19 Feb 2013, 09:15
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