Last visit was: 20 Nov 2025, 06:23 It is currently 20 Nov 2025, 06:23
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3   
User avatar
Chiranjeevee
User avatar
Joined: 28 Jan 2013
Last visit: 12 Aug 2014
Posts: 23
Own Kudos:
40
Given Kudos: 20
Posts: 23
Kudos: 40
If the graph of y = x^2 + ax + b passes through the points 21 Oct 2013, 02:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Sam1
This is how I am solving it. Will appreciate your help to tell me where I am going wrong.
since m and n are the two points through which the graph passes they will both satisfy the equation.
Hence
0=m^2+am+b
0=n^2+an+b
1-2
gives
0=n^2-m^2 +a (n-m)
taking n-m common
(n-m) (n+m+a)=0
hence (n-m)=0 or (n+m+a)=0
n=m; (n+m)=-a
looking at statement a
a^2-4=4b
a will be something in terms of b but we will not the exact value of a.
Please tell me where am I making a mistake
Show more


n and m are the roots of the equation, so

(n+m)=-a
and nm=b
and we need to find n-m. so just solve the equation to get (n-m) term.

(n-m)^2= (n+m)^2 - 4mn

put above values

(n-m)^2= (n+m)^2 - 4mn = a^2-4b = 4 (as per the given equation, i)

hence n-m = 2
User avatar
Sam1
User avatar
Joined: 18 Sep 2013
Last visit: 16 Mar 2014
Posts: 21
Given Kudos: 1
Posts: 21
Kudos: 0
If the graph of y = x^2 + ax + b passes through the points 21 Oct 2013, 05:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Chiranjeevee
Sam1
This is how I am solving it. Will appreciate your help to tell me where I am going wrong.
since m and n are the two points through which the graph passes they will both satisfy the equation.
Hence
0=m^2+am+b
0=n^2+an+b
1-2
gives
0=n^2-m^2 +a (n-m)
taking n-m common
(n-m) (n+m+a)=0
hence (n-m)=0 or (n+m+a)=0
n=m; (n+m)=-a
looking at statement a
a^2-4=4b
a will be something in terms of b but we will not the exact value of a.
Please tell me where am I making a mistake


n and m are the roots of the equation, so

(n+m)=-a
and nm=b
and we need to find n-m. so just solve the equation to get (n-m) term.

(n-m)^2= (n+m)^2 - 4mn

put above values

(n-m)^2= (n+m)^2 - 4mn = a^2-4b = 4 (as per the given equation, i)

hence n-m = 2
Show more

Hi thank you for your reply. While I understand the part with the sum of roots. I am still a little confused of how are we ignoring the b term. Are we equating (n-m)^2 to a^2?
User avatar
russ9
User avatar
Joined: 15 Aug 2013
Last visit: 20 Apr 2015
Posts: 174
Own Kudos:
400
Given Kudos: 23
Posts: 174
Kudos: 400
If the graph of y = x^2 + ax + b passes through the points 18 May 2014, 12:19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
gabrieldoria
Sorry I´ve got one last question after reading the whole thread.

Wouldn´t option 1 be insufficient too, because the SQ ROOT of 4 be +- 2? That would give two possible values for n-m.

No, \(\sqrt{4}=2\), not +2 or -2.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{25}=5\), NOT +5 or -5.

In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Hope it helps.
Show more

Hi Bunuel,

This is news to me. Is this valid on ALL DS and PS problems or just one subset? Does this hold true for inequalities as well?

Also, can you suggest similar quadratics that use vieta's theorem and require us to relate roots to quadratics?

Thanks!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Nov 2025
Posts: 105,416
Own Kudos:
778,504
 [1]
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,416
Kudos: 778,504
 [1]
If the graph of y = x^2 + ax + b passes through the points 19 May 2014, 03:12
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
russ9
Bunuel
gabrieldoria
Sorry I´ve got one last question after reading the whole thread.

Wouldn´t option 1 be insufficient too, because the SQ ROOT of 4 be +- 2? That would give two possible values for n-m.

No, \(\sqrt{4}=2\), not +2 or -2.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{25}=5\), NOT +5 or -5.

In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Hope it helps.

Hi Bunuel,

This is news to me. Is this valid on ALL DS and PS problems or just one subset? Does this hold true for inequalities as well?

Also, can you suggest similar quadratics that use vieta's theorem and require us to relate roots to quadratics?

Thanks!
Show more

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.
User avatar
gmatacequants
User avatar
Joined: 13 May 2014
Last visit: 03 Sep 2019
Posts: 26
Own Kudos:
258
Given Kudos: 1
Concentration: General Management, Strategy
Posts: 26
Kudos: 258
If the graph of y = x^2 + ax + b passes through the points 19 May 2014, 03:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Sam1
This is how I am solving it. Will appreciate your help to tell me where I am going wrong.
since m and n are the two points through which the graph passes they will both satisfy the equation.
Hence
0=m^2+am+b
0=n^2+an+b
1-2
gives
0=n^2-m^2 +a (n-m)
taking n-m common
(n-m) (n+m+a)=0
hence (n-m)=0 or (n+m+a)=0
n=m; (n+m)=-a
looking at statement a
a^2-4=4b
a will be something in terms of b but we will not the exact value of a.
Please tell me where am I making a mistake
Show more

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
User avatar
pretzel
User avatar
Joined: 04 Jan 2014
Last visit: 28 Oct 2014
Posts: 80
Own Kudos:
68
Given Kudos: 24
Posts: 80
Kudos: 68
If the graph of y = x^2 + ax + b passes through the points 17 Jun 2014, 21:53
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mbaiseasy
Given \(F(x) = Ax^2 + Bx + C\)
We could get the difference of roots with a formula:
\(x1-x2=\sqrt{{\frac{b^2-4ac}{a^2}}}\) with \(x1>x2\)

Show more

Should the denominator be \(a^2\) or \(2a\)?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Nov 2025
Posts: 105,416
Own Kudos:
778,504
 [1]
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,416
Kudos: 778,504
 [1]
If the graph of y = x^2 + ax + b passes through the points 18 Jun 2014, 01:32
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pretzel
mbaiseasy
Given \(F(x) = Ax^2 + Bx + C\)
We could get the difference of roots with a formula:
\(x1-x2=\sqrt{{\frac{b^2-4ac}{a^2}}}\) with \(x1>x2\)


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

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.
User avatar
Reaper
User avatar
Joined: 31 May 2013
Last visit: 28 Aug 2018
Posts: 10
Own Kudos:
4
Given Kudos: 31
Products:
My Reviews
Posts: 10
Kudos: 4
If the graph of y = x^2 + ax + b passes through the points 20 Oct 2014, 06:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
avatar
saadis87
avatar
Joined: 13 Dec 2013
Last visit: 19 Apr 2016
Posts: 30
Own Kudos:
24
Given Kudos: 21
Schools: Booth '18 Ross '18 CBS '18 Rutgers '18 Smeal Penn'18 Sauder '18 Rotman '17 Stern '18 Fisher Pepperdine '18
GPA: 2.71
Schools: Booth '18 Ross '18 CBS '18 Rutgers '18 Smeal Penn'18 Sauder '18 Rotman '17 Stern '18 Fisher Pepperdine '18
Posts: 30
Kudos: 24
If the graph of y = x^2 + ax + b passes through the points 05 Dec 2014, 10:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mbaiseasy
Given \(F(x) = Ax^2 + Bx + C\)
We could get the difference of roots with a formula:
\(x1-x2=\sqrt{{\frac{b^2-4ac}{a^2}}}\) with \(x1>x2\)

Solution:
Apply that formula to the problem:
\(y=x^2+ax+b\) that passes (m,0) and (n,0) with n > m

\(n-m = \sqrt{\frac{a^2-4(1)(b)}{(1)^2}}=\sqrt{a^2-4b}\)

Statement (1) gives us \(\sqrt{a^2-4b}=\sqrt{4}=2\) SUFFICIENT.
Statement (2) gives us b = 0. Thus, INSUFFICIENT.

Formula for determining the SUM, PRODUCT and DIFFERENCE of roots of \(F(x) = Ax^2 + Bx + C\):
https://burnoutorbreathe.blogspot.com/2012/12/sum-and-product-of-roots-of-fx.html
https://burnoutorbreathe.blogspot.com/2012/12/algebra-difference-of-roots-of-fx.html
Show more


Now why didnt i read this before I read the question lol
avatar
anupamadw
avatar
Joined: 31 Jul 2014
Last visit: 29 Jun 2016
Posts: 106
Own Kudos:
139
Given Kudos: 373
GMAT 1: 630 Q48 V29
GMAT 1: 630 Q48 V29
Posts: 106
Kudos: 139
If the graph of y = x^2 + ax + b passes through the points 08 Dec 2014, 03:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
EvaJager
aeros232
If the graph of y = x^2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a^2 – 4

(2) b = 0

Show Spoiler
A
Show more

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 ?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Nov 2025
Posts: 105,416
Own Kudos:
778,504
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,416
Kudos: 778,504
If the graph of y = x^2 + ax + b passes through the points 08 Dec 2014, 04:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
anupamadw
EvaJager
aeros232
If the graph of y = x^2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a^2 – 4

(2) b = 0

Show Spoiler
A

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 ?
Show more

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 \(x=\sqrt{9}=3\).
avatar
target760gmat
avatar
Joined: 15 Jun 2016
Last visit: 06 Dec 2016
Posts: 42
Own Kudos:
44
Given Kudos: 42
Location: India
Concentration: Technology, Strategy
Schools: ISB '18 SPJ PGPM"17 NUS '18
GMAT 1: 730 Q50 V39
Schools: ISB '18 SPJ PGPM"17 NUS '18
GMAT 1: 730 Q50 V39
Posts: 42
Kudos: 44
If the graph of y = x^2 + ax + b passes through the points 09 Jul 2016, 23:36
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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 ...
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Nov 2025
Posts: 105,416
Own Kudos:
778,504
 [1]
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,416
Kudos: 778,504
 [1]
If the graph of y = x^2 + ax + b passes through the points 09 Jul 2016, 23:43
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
target760gmat
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 ...
Show more

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.
avatar
target760gmat
avatar
Joined: 15 Jun 2016
Last visit: 06 Dec 2016
Posts: 42
Own Kudos:
44
Given Kudos: 42
Location: India
Concentration: Technology, Strategy
Schools: ISB '18 SPJ PGPM"17 NUS '18
GMAT 1: 730 Q50 V39
Schools: ISB '18 SPJ PGPM"17 NUS '18
GMAT 1: 730 Q50 V39
Posts: 42
Kudos: 44
If the graph of y = x^2 + ax + b passes through the points 09 Jul 2016, 23:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
thanks Bunuel for clarifying my doubt... :-D

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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Nov 2025
Posts: 105,416
Own Kudos:
778,504
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,416
Kudos: 778,504
If the graph of y = x^2 + ax + b passes through the points 09 Jul 2016, 23:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
target760gmat
thanks Bunuel for clarifying my doubt... :-D

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
Show more

What do you mean by "such questions"? Or by "other than equations"?
avatar
target760gmat
avatar
Joined: 15 Jun 2016
Last visit: 06 Dec 2016
Posts: 42
Own Kudos:
44
Given Kudos: 42
Location: India
Concentration: Technology, Strategy
Schools: ISB '18 SPJ PGPM"17 NUS '18
GMAT 1: 730 Q50 V39
Schools: ISB '18 SPJ PGPM"17 NUS '18
GMAT 1: 730 Q50 V39
Posts: 42
Kudos: 44
If the graph of y = x^2 + ax + b passes through the points 10 Jul 2016, 00:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Yes.. Other than equations where such type of questions are asked (only if you have the link readily).. thanks
User avatar
KARISHMA315
User avatar
Joined: 09 May 2016
Last visit: 01 Jan 2024
Posts: 52
Own Kudos:
55
Given Kudos: 12
Location: United States (NY)
Concentration: Finance, General Management
Schools: Wharton EMBA'26 (A) Yale (A)
GMAT 1: 710 Q49 V38
WE:Information Technology (Consulting)
Schools: Wharton EMBA'26 (A) Yale (A)
GMAT 1: 710 Q49 V38
Posts: 52
Kudos: 55
If the graph of y = x^2 + ax + b passes through the points 26 Mar 2017, 03:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
EvaJager
aeros232
If the graph of y = x^2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a^2 – 4

(2) b = 0

Show Spoiler
A

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 \(n-m=\sqrt{4}=2.\)
(2) The quadratic equation becomes \(x^2+ax=0\) or \(x(x+a)=0.\)
Not sufficient, as one of the roots is \(0\) and the other one is \(-a\), and we have no information about \(a\).

Answer A.
Show more

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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Nov 2025
Posts: 105,416
Own Kudos:
778,504
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,416
Kudos: 778,504
If the graph of y = x^2 + ax + b passes through the points 26 Mar 2017, 05:08
Kudos
Add Kudos
Bookmarks
Bookmark this Post
KARISHMA315
EvaJager
aeros232
If the graph of y = x^2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a^2 – 4

(2) b = 0

Show Spoiler
A

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 \(n-m=\sqrt{4}=2.\)
(2) The quadratic equation becomes \(x^2+ax=0\) or \(x(x+a)=0.\)
Not sufficient, as one of the roots is \(0\) and the other one is \(-a\), and we have no information about \(a\).

Answer A.

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
Show more

You have to go through the basics before attempting questions. Check here: https://gmatclub.com/forum/coordinate-ge ... 87652.html
avatar
chessgiants
avatar
Joined: 13 Apr 2017
Last visit: 23 Jan 2018
Posts: 33
Own Kudos:
5
Given Kudos: 2
Posts: 33
Kudos: 5
If the graph of y = x^2 + ax + b passes through the points 08 May 2017, 00:40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
the problem is interesting, but it takes me more than 2 min to find the correct answer
User avatar
mba20202022
User avatar
Joined: 04 Apr 2018
Last visit: 02 Jan 2022
Posts: 52
Own Kudos:
23
Given Kudos: 405
Posts: 52
Kudos: 23
If the graph of y = x^2 + ax + b passes through the points 01 Dec 2018, 03:43
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
   1   2   3   
Moderators:
Bunuel
Math Expert
105416 posts
kingbucky
496 posts