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

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



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

Could someone tell me if it is an important, really importante question, or negligible ???

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:
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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.

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

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 = B

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.

The answer is “B”.

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.

Video solution from Quant Reasoning:
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Solution:

Question Stem Analysis:

We need to determine whether rs < 0 given that r and s are the roots of the equation x^2 + bx + c = 0. Recall that for any quadratic equation ax^2 + bx + c = 0 where a ≠ 0, the product of the two roots is c/a. Since here a = 1 and r and s are the roots, rs = c/1 = c. Therefore, if we know c is negative, then rs < 0.

Statement One Alone:

Since we don’t know anything about the value c, statement one is not sufficient.

Statement Two Alone:

Since c < 0, rs < 0. Statement two is sufficient.

Answer: B

Answer: Option B

Video solution by GMATinsight



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Whenever I encounter a question on Quadratic equations, the first thing I do is to compare the given equation to the standard form of a Quadratic equation viz., a\(x^2\) + bx + c = 0 and write down the values of a, b and c for the given equation.

The given equation is \(x^2\) + bx + c = 0. Comparing with the standard form, a = 1, b = b and c = c.
In its standard form, the sum of the roots of a quadratic equation = -b/a and product of the roots = c/a

Therefore, for the given equation, sum of roots = r + s = -b and product of roots = rs = c, since the question clearly says that r and s are the roots of the equation.
It turns out that rs is the product of the roots of the equation. Therefore, we can rephrase the question stem now to make it “Is c<0?”

From statement I alone, b<0. This is insufficient to say if c<0 since the sum of two numbers could be negative but their product could be ≥ 0.
Statement I alone is insufficient. Answer options A and D can be eliminated.

From statement II alone, c<0. This answers the question directly with a YES.
Statement II alone is sufficient. Answer options C and E can be eliminated.

The correct answer option is B.

Hope that helps!
Aravind BT

Hi Scott, It would also be good to have this rule about sum and product of roots in the TTP course.

avigutman thank you for this.

I am a bit confused as to how you knew that "rs" was = to C.

Why isn't -rx + rs = to C for instance?

Is it because you know that for "bx" must have an x and "r" and "s" are coefficients? Am I using the right terminology?

Thank you again.

Yes, that's exactly right, woohoo921. In school we called that "combining like terms."
So we're going from here:
x^2 - xs - rx + rs = 0
To here:
x^2 - rx - sx + rs = 0
And end up here:
x^2 - (r+s)x + rs = 0

Hi BrentGMATPrepNow, in the green highlighted line, why is r + s = -b in x² + bx + c = 0, then r + s = -b, and rs = c as there's no any negative sign in the equation? Have I missed something here? Thanks Brent.

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