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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]

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21 Oct 2012, 02:08

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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]

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13 Nov 2012, 08:34

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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.

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

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5

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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.

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

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.

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.

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

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

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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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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

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). _________________

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

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13 Apr 2014, 10:21

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!

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 (\(1*x^2+bx+c=0\)).

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

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16 Aug 2014, 08:19

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!
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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 Nov 2014, 12:49

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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.