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If r and s are the roots of the equation x^2 + bx + 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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New post 20 Nov 2014, 10:16
Hi,

can someone please confirm that the "roots" this question talks about is something like this: (x-xyz) * (x+asd) ? If that is the case, then this question becomes at least understandable for me...
Until now, a "root" was always something like sqrt(2) for me (here, a square root).

Thanks and best regards
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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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New post 14 Dec 2014, 09:29
in the quadratic equation C will be the two roots multiplied together...to know if rs < 0 we just need to know if either R or S is negative.

1.) knowing the value of B is negative doesn't really get us anywhere. R and S could both be negative or one could be negative

2.) we know that c=rs...c < 0 so lets look at what roots will produce C < 0

(x-r)(x-s) means C will be positive
(x+r)(x+s) means C will be positive

Therefore either R or S is negative as shown below

(x-r)(x+s) means C will be negative
(x+r)(x-s) means c will be negative

Therefore we can determine if rs < 0 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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New post 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
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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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New post 23 May 2016, 06:45
1
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.


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 ???


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.


I don't understand why have you considered the roots (x-r) and (x-s). Couldn't it even be (x + r) & (x + s)
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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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New post 23 May 2016, 06:56
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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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If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

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New post 08 Jul 2016, 12:29
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.
Can this question nevertheless be solved based on the above determinant formula?



I come a bit late but having solved it with the over-mentioned formula I feel it's worth sharing.

So, here we go:
Unless I made some mistake the generic formula for quadratic equations works like a charm.

The formula to solve quadratic equations is (given \(ax^2 +bx +c=0\)):
\(\frac{-b±√((b^2)-4ac)}{2}\)

So if we do:
\(\frac{-b+√((b^2)-4ac)}{2}\)*\(\frac{-b-√((b^2)-4ac)}{2}\)
we have a difference of squares resulting in:

\(\frac{((b)^2-(b^2-4ac))}{4}\) -> which simplified gives us (remember that in this case a=1): \(\frac{(+4c))}{4}\) -> c

And we now know that all we need to make r*s<0 is c<0, and so statement 2 answers is what we are looking for.

Cheers,
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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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New post 16 Aug 2016, 15:59
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asveaass wrote:
If r and s are the roots of the equation x² + bx + c = 0, where b and c are constants, is rs < 0 ?

(1) b < 0
(2) c < 0


Let's first examine the relationship between the roots of an equation and the given equation. Here are some examples:

Example #1: x² - 5x + 6 = 0
We can rewrite this as x² + (-5x) + 6 = 0 [to make it look like the given equation x² + bx + c = 0]
So, b = -5 and c = 6
To solve the equation, we'll factor to get: (x - 3)(x - 2) = 0
So, the ROOTS of the equation are x = 2 and x = 3
NOTICE that the sum of the roots equals -b, and notice that the product of the roots = c

Example #2: x² + 6x - 7 = 0
We can rewrite this as x² + 6x + (-7) = 0 [to make it look like the given equation x² + bx + c = 0]
So, b = 6 and c = -7
To solve the equation, we'll factor to get: (x + 7)(x - 1) = 0
So, the ROOTS of the equation are x = -7 and x = 1
NOTICE that the sum of the roots equals -b, and notice that the product of the roots = c

We could keep going with more examples, but the big takeaway is as follows:
If r and s are the roots of the equation x² + bx + c = 0, then r + s = -b, and rs = c

Okay, now onto the question....

Target question: Is rs < 0?

Given: r and s are the roots of the equation x² + bx + c = 0

Statement 1: b < 0
This means that b is NEGATIVE, which also means that -b is POSITIVE
From our conclusions above, we saw that r + s = -b
So, we can now conclude that r + s = some POSITIVE VALUE.
Is this enough info to determine whether rs < 0?
NO.
Consider these two conflicting cases:
Case a: r = -1 and s = 2 (here r + s = some positive value), in which case rs < 0
Case b: r = 1 and s = 2 (here r + s = some positive value), in which case rs > 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: c < 0
From our conclusions above, we saw that rs = c
Now, statement 2 tells us that c is negative.
So, it MUST be the case that rs < 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer =

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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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New post 31 Aug 2016, 16:34
Top Contributor
asveaass 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 ?

(1) b < 0
(2) c < 0


Let's first examine the relationship between the roots of an equation and the given equation. Here are some examples:

Example #1: x² - 5x + 6 = 0
We can rewrite this as x² + (-5x) + 6 = 0 [to make it look like the given equation x² + bx + c = 0]
So, b = -5 and c = 6
To solve the equation, we'll factor to get: (x - 3)(x - 2) = 0
So, the ROOTS of the equation are x = 2 and x = 3
NOTICE that the sum of the roots equals -b, and notice that the product of the roots = c

Example #2: x² + 6x - 7 = 0
We can rewrite this as x² + 6x + (-7) = 0 [to make it look like the given equation x² + bx + c = 0]
So, b = 6 and c = -7
To solve the equation, we'll factor to get: (x + 7)(x - 1) = 0
So, the ROOTS of the equation are x = -7 and x = 1
NOTICE that the sum of the roots equals -b, and notice that the product of the roots = c

We could keep going with more examples, but the big takeaway is as follows:
If r and s are the roots of the equation x² + bx + c = 0, then r + s = -b, and rs = c

Okay, now onto the question....

Target question: Is rs < 0?

Given: r and s are the roots of the equation x² + bx + c = 0

Statement 1: b < 0
This means that b is NEGATIVE, which also means that -b is POSITIVE
From our conclusions above, we saw that r + s = -b
So, we can now conclude that r + s = some POSITIVE VALUE.
Is this enough info to determine whether rs < 0?
NO.
Consider these two conflicting cases:
Case a: r = -1 and s = 2 (here r + s = some positive value), in which case rs < 0
Case b: r = 1 and s = 2 (here r + s = some positive value), in which case rs > 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: c < 0
From our conclusions above, we saw that rs = c
Now, statement 2 tells us that c is negative.
So, it MUST be the case that rs < 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer =

Cheers,
Brent
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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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New post 10 Nov 2016, 13:38
Answer:B
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
1- b= -(r+s); b<0 ⇒ -(r+s)<0 ⇒ r+s>0 then r&s could both be positive rs>0 or one positive and another negative rs<0 -- NOT SUFFICIENT
2- c is equal to rs; c<0 ⇒ rs<0 -- SUFFICIENT
e.g. x^2+2x-8=(x-2)(x+4) -- c=-8, r=2, and s=-4 -- rs=-8=c
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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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New post 19 Feb 2017, 09:26
Give equation is quadratic. Since r and s are the roots of the equation (-r)*(-s)=c and (-r)+(-s)=b

To find out if rs is negative, we need to know the signs of each multiplier or we need to know the sign of the product.

1) Statement one tells us that b=-r-s is negative. This could be true if both r and s are positive, or if the multiplier with larger absolute value is negative, and the other multiplier is positive. Since there are two possible scenarios, this statement is insufficient.
2) Statement two tells us that c=(-r)*(-s) is negative, and this is just the information that we were looking for. Since c is negative, then both r and s are negative or they both are positive. In either case, the product of the two is positive.

Statement 2 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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New post 28 Jul 2017, 01:46
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

My 2 cents.
From the question, we know that (x-r)(x-s).
If we solve this, we can x^2-rx-sx+rs, which is x^s+x(-r-s)+rs
And this equals x^2 +bx+c. So...
b=-r-s and C=rs.
B says C is negative, then rs = negative, thus less than O.

Hence 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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