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

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

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Updated on: 07 Dec 2012, 09:05
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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.
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If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

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23 Oct 2012, 08:07
35
96
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.
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b  [#permalink]

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14 Nov 2012, 10:41
22
17
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.
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b  [#permalink]

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20 Oct 2012, 12:21
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3
Product of roots = c/a = c (in this case)

So question is basically asking if c<0.

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

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20 Jan 2013, 03:21
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).
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Finance your Student loan through SoFi and get $100 referral bonus : Click here Senior Manager Joined: 15 Jun 2010 Posts: 270 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 21 Oct 2012, 02:08 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. 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 Math Expert Joined: 02 Sep 2009 Posts: 60460 Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink] ### Show Tags 14 Apr 2014, 02:07 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. _________________ e-GMAT Representative Joined: 04 Jan 2015 Posts: 3209 Re: If r and s are the roots of the equation x^2 + bx + c = 0 [#permalink] ### Show Tags 22 Apr 2015, 00:57 2 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 _________________ Board of Directors Joined: 01 Sep 2010 Posts: 3412 Re: If r and s are the roots of the equation x^2+bx+c=0, where b [#permalink] ### Show Tags 13 Nov 2012, 08:34 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. 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 ??? _________________ VP Joined: 02 Jul 2012 Posts: 1096 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 14 Feb 2013, 21:47 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 = $$\frac{c}{a}$$ 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]

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

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23 May 2016, 06:56
1
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...
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Re: If r and s are the roots of the equation x^2+bx+c=0, where b  [#permalink]

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21 Oct 2012, 05:32
asveaass wrote:

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

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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.

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

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14 Jan 2013, 21:02
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
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Re: If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

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

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14 Feb 2013, 18:05
At tthe time this question drove me insane but indeed is fairly stupid

if you need something else ask ;9
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Re: If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

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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  [#permalink]

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04 Mar 2014, 03:53
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
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Re: If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

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04 Mar 2014, 11:56

(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
Re: If r and s are the roots of the equation x^2 + bx + c = 0   [#permalink] 04 Mar 2014, 11:56

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