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Intern  Joined: 29 Aug 2013
Posts: 11
Re: If r and s are the roots of the equation x^2 + bx + c = 0  [#permalink]

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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
Senior Manager  Joined: 13 Jun 2013
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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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mott wrote:
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

you're right. root here means factor of the given quadratic equation.
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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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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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Joined: 04 Jan 2015
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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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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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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.

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.

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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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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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,
Den
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Joined: 12 Sep 2015
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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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Top Contributor
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

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

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Top Contributor
Oops - duplicate response!

Cheers,
Brent
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Originally posted by GMATPrepNow on 31 Aug 2016, 16:34.
Last edited by GMATPrepNow on 13 Mar 2019, 13:13, edited 1 time in total.
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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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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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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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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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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 $$rs=\frac{c}{1}=c$$. So, we are basically asked whether $$c<0$$.

(1) b<0. Not sufficient.
(2) c<0. Directly answers the question. Sufficient.

Hope it helps.

Like really? Vietes theorem? This was not the best expl for the B man ?
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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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Let's talk strategy here. Many explanations of Quantitative questions focus blindly on the math, but remember: the GMAT is a critical-thinking test. For those of you studying for the GMAT, you will want to internalize strategies that actually minimize the amount of math that needs to be done, making it easier to manage your time. The tactics I will show you here will be useful for numerous questions, not just this one. My solution is going to walk through not just what the answer is, but how to strategically think about it. Ready? Here is the full “GMAT Jujitsu” for this question:

First, this problem requires you to understand positive/negative rules. When adding two numbers together,
(+) + (+) = (+)
(-) + (-) = (-)
(+) + (+) = ???
When multiplying two numbers together,
(+) * (+) = (+)
(+) * (-) = (-)
(-) * (-) = (+)

Next, let’s talk terminology. When a problem refers to the “roots” of a quadratic equation containing $$x$$, it is talking about the possible numerical solutions for $$x$$. In this context, the term “root” is completely synonymous with the term “solution.” The solutions of quadratics can often be calculated by breaking down the quadratic into its linear components such that

$$x^2+bx+c=0$$

becomes $$(x+\underline{\hspace{2em}})(x+ \underline{\hspace{2em}})=0$$

You fill in the blanks here by looking at the pairs of factors of $$c$$ (called “complementary factors”) that when multiplied together equals $$c$$, but when added together they equal $$b$$. Of course, if you need to actually solve for the values, it is important to realize that the solutions (or roots) of the equation are going to be the negatives of these factors. (Think about it, if $$(x+4)=0$$, then $$x$$ must be $$-4$$.)

Now that we have the basic rules, down, let’s focus on tactics. This is a “Yes/No” question – a very common structure for Data Sufficiency problems. The fundamental trap for problems like these is to bait you into thinking that you actually need to solve for every value. You don’t. As soon as you have enough information to conclude that a statement is either sufficient or insufficient, you can move on. For “Yes/No” questions, if you can think of two situations (or two variable inputs) that are consistent with all of the problem’s constraints but come up with different answers to the question, you know a statement is insufficient. In my classes, I call this strategy “Play Both Sides.” The problem asks us, “is $$rs < 0$$?” – in other words, “when you multiply the two solutions to a quadratic equation, do you get a negative number?” We can start to anticipate how we are going to “Play Both Sides.” Using the positive/negative rules we showed above, the answer to this question is “Yes” if one of the solutions is positive and the other is negative. The answer is “No” if either both solutions are positive or both solutions are negative.

Statement #1 tells us that $$b$$ is negative. There are a couple of ways this could happen: (1) the complementary factors of $$c$$ could be both negative, in which case $$r$$ and $$s$$ are both positive, or (2) one complementary factor of $$c$$ is positive and the other is negative (so long as the negative factor is larger in magnitude than the positive factor.) This would mean that we would have one negative solution and one positive one. Since the first possibility gives us a “No” answer to the question while the second option gives us a “Yes” answer, Statement #1 is not sufficient.

Statement #2 tells us that $$c$$ is negative. There is only one possible way this could happen: one complementary factor of $$c$$ is positive and the other is negative. You can only get a negative product if one factor is positive and one is negative. The roots would mirror this. So, Statement #2 can only give us a “Yes” answer to the Yes/No question. Since we have only one answer, it is sufficient.

Now, let’s look back at this problem through the lens of strategy. This question can teach us patterns seen throughout the GMAT. Notice that this problem is much more about logic and critical-thinking than it is about math. With “Yes/No” questions involving positive/negative rules, a great tactic that you can often use is to focus conceptually on the +/- possibilities, using very basic rules. If multiple answers are possible using those rules, you can “Play Both Sides” and disprove sufficiency. On the other hand, if there is only one answer to the Yes/No question, then you proved sufficiency. If you can analyze the problems conceptually, the actual math involved is often minimal, and you don’t even need to solve for specific values. And that is how you think like the GMAT.
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