If the graph of y = x^2 + ax + b passes through the points

Yes, this is true for all GMAT questions.

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
if-r-and-s-are-the-roots-of-the-equation-x-2-bx-c-141018.html

Hope this helps.

Hi Sam1,
Basically we need to find the value of n-m, the difference between the roots.
What you have done is, showed the sum of the roots , n+m = - a ---(1) which you have deduced right.
NOW,
multiplying the above equation by n yields:
n^2+m*n = -a*n
or n^2 +m*n + a*n =0
or m*n - b = 0 ---- (as n^2+a*n+b =0 ,n being one of the roots)
or m*n = b ----------(2)

Now, (n-m)^2 = n^2 -2*m*n + m^2
= (n+m)^2 - 4*m*n

using eqn(1) and eqn(2), we have
(n-m)^2 = a^2 - 4*b
or (n-m) = sq.root(a^2 - 4*b)

Now from answer choices,using option 1,we can find n-m = 2
however,using option 2, we can not find out n-m.
Hence, A. Hope this helps.

Press kudos if you wish to appreciate

Should the denominator be \(a^2\) or \(2a\)?

It's correct as it is:

\(x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\);

\(x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}\);

\(x_1-x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}-\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{\sqrt{b^2-4ac}}{a}=\frac{\sqrt{b^2-4ac}}{\sqrt{a^2}}\).

Hope it's clear.

From 1, obtain 2 quadratic equations for \(y=x^2 + ax +b\) by substituting (m,0) and (n,0)

\(m^2 + am + b\) = 0
\(n^2 + am + b\) = 0

which gives

n = (-a+\(\sqrt{a^2-4b}\))/2a ; (-b-\(\sqrt{a^2-4b}\))/2a
m = (-a+\(\sqrt{a^2-4b}\))/2a ; (-b-\(\sqrt{a^2-4b}\))/2a

since n - m > 0,

n-m = \(\sqrt{a^2-4b}\)

substitute from (1) \(a^2-4b\) = 4

n-m = \(\sqrt{4}\)
n-m =2
Sufficient

Now why didnt i read this before I read the question lol

We are given that \(m\) and \(n\) are the roots of the quadratic equation \(x^2+ax+b=0.\)
They are given by the formula \(x_{1,2}=\frac{-a\pm\sqrt{a^2-4b}}{2}\).
The larger root \(n\) is that with a plus in front of the square root, therefore \(n-m=\frac{a}{2}+\frac{\sqrt{a^2-4b}}{2}-(\frac{a}{2}-\frac{\sqrt{a^2-4b}}{2})=\sqrt{a^2-4b}\).

(1) Sufficient, because \(a^2-4b=4\), so \(n-m=\sqrt{4}=2.\)
here we dont have to consider +2 and -2 ?

Because \(\sqrt{4}=2\) ONLY, not +/-2.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

A doubt - In question all the provided information is in terms of variable x, constants a,b,c etc.
If our objective is to able to find a solution - then by taking only (2) gives us the answer of 'a' right ?
I do agree that we don't know the value of a, however according to question a is provided a constant and it will always be it..
Pls clarify ...

When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.

thanks Bunuel for clarifying my doubt...

Are there any other topics other than equations in which such questions could be asked? Or if you have the link it would be helpful.. thanks

What do you mean by "such questions"? Or by "other than equations"?

Yes.. Other than equations where such type of questions are asked (only if you have the link readily).. thanks

I have a very silly doubt but pls help. What in the question made you think that m and n are roots ,why cant they be a point (m,n) from which the graph passes

You have to go through the basics before attempting questions. Check here: https://gmatclub.com/forum/coordinate-ge ... 87652.html

the problem is interesting, but it takes me more than 2 min to find the correct answer

MGMAT official solution.
could someone please explain their explanation for the second statement? (in red below)
Bunuel


"Before looking at the statements, make sure you understand what the question is asking. In particular, what is the connection between a, b, m, and n?

The function y = x^2 + ax + b is quadratic, so it can’t intersect the x-axis in more than two places. Since (m, 0) and (n, 0) are distinct x-intercepts, they must be the only two x-intercepts of the parabola. Therefore, the equation of the parabola can be written as y = (x – m)(x – n), which can be expanded into y = x^2 – mx – nx + mn, or y = x^2 + (­–m – n)x + mn. Since this equation must be equal to the equation y = x^2 + ax + b, it follows that a = –m – n and b = mn.

(1) SUFFICIENT: We have equations that relate a and b to m and n. Replace a with (−m – n) and b with (mn):

4(mn) = (–m – n)2 – 4
4mn = m2 + 2mn + n2 – 4
4 = m2 – 2mn + n2
4 = (m – n)2
m – n = 2 or –2

Since the problem specifies that m < n, the first of these is impossible. Therefore, m – n = –2, or n – m = 2.

We can also deal with this statement by using smart numbers.

Solve the statement by dividing by 4, giving b = a^2/4 – 1. Then substitute values for a, solve for b, and then find m and n by solving the resulting quadratic.

· If a = 2, then b = 4/4 – 1 = 0. Therefore, the equation is y = x^2 + 2x, which factors to y = x(x + 2). The x-intercepts of this function are m = –2 and n = 0, so n – m = 2.

· If a = 4, then b = 16/4 – 1 = 3. Therefore, the equation is y = x^2 + 4x + 3, which factors to y = (x + 1)(x + 3). The x-intercepts of this function are m = –3 and n = –1, so n – m = 2.

· If a = 6, then b = 36/4 – 1 = 8. Therefore, the equation is y = x^2 + 6x + 8, which factors to y = (x + 2)(x + 4). The x-intercepts of this function are m = –4 and n = –2, so n – m = 2.

Try more cases if necessary (you may want to try a = 0, or a = negative); you’ll get n – m = 2 every time.

(2) INSUFFICIENT: If b = 0, then the equation is y = x2 + ax, where a is an unspecified constant. This equation factors to y = x(x + a) and so has x-intercepts –a and 0. Therefore, n – m = 0 – (–a) = a. Since a is unspecified, there are many possible values.

We can also deal with this statement by using smart numbers.

We know b = 0; pick different values for a.

· If a = 1, then the equation is y = x^2 + x, which factors to y = x(x + 1). The x-intercepts of this equation are –1 and 0, so n – m = 0 – (–1) = 1.

· If a = 2, then the equation is y = x^2 + 2x, which factors to y = x(x + 2). The x-intercepts of this equation are –2 and 0, so n – m = 0 – (–2) = 2.

We have found two different values for n – m, so this statement is insufficient.

The correct answer is A.