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.
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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.

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

Relevant post:

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



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

Could someone tell me if it is an important, really importante question, or negligible ???
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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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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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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
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For any equation, \(ax^2 + bx + c = 0\)

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

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

gmat dose not require us to remember formular

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

so, c=rs.

B

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

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

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



Hope this helps.
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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!

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.