Last visit was: 19 Nov 2025, 04:02 It is currently 19 Nov 2025, 04:02
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 
avatar
Hanzel101
avatar
Joined: 15 Jul 2016
Last visit: 23 Feb 2020
Posts: 72
Own Kudos:
92
Given Kudos: 190
Location: India
Schools: Oxford "21 (A) Cambridge "21 (A$)
GMAT 1: 690 Q48 V36
Schools: Oxford "21 (A) Cambridge "21 (A$)
GMAT 1: 690 Q48 V36
Posts: 72
Kudos: 92
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 20 Aug 2019, 20:02
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi All,

can somebody please help me understand this? Here, when we get (x-a)(x-b) = 0, then why aren't the roots x=a and x=b ?

Then statement 1 will be sufficient ?

Please help!

VeritasKarishma Bunuel egmat chetan2u
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,379
Own Kudos:
778,194
Given Kudos: 99,977
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: 105,379
Kudos: 778,194
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 20 Aug 2019, 21:20
Kudos
Add Kudos
Bookmarks
Bookmark this Post
578vishnu
Hi All,

can somebody please help me understand this? Here, when we get (x-a)(x-b) = 0, then why aren't the roots x=a and x=b ?

Then statement 1 will be sufficient ?

Please help!

VeritasKarishma Bunuel egmat chetan2u
Show more

This is a value question, which means that for a statement(s) to be sufficient you need to get single numerical value of the intersect points.
avatar
Hanzel101
avatar
Joined: 15 Jul 2016
Last visit: 23 Feb 2020
Posts: 72
Own Kudos:
92
Given Kudos: 190
Location: India
Schools: Oxford "21 (A) Cambridge "21 (A$)
GMAT 1: 690 Q48 V36
Schools: Oxford "21 (A) Cambridge "21 (A$)
GMAT 1: 690 Q48 V36
Posts: 72
Kudos: 92
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 20 Aug 2019, 21:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
578vishnu
Hi All,

can somebody please help me understand this? Here, when we get (x-a)(x-b) = 0, then why aren't the roots x=a and x=b ?

Then statement 1 will be sufficient ?

Please help!

VeritasKarishma Bunuel egmat chetan2u

This is a value question, which means that for a statement(s) to be sufficient you need to get single numerical value of the intersect points.
Show more

Hi Bunuel

I agree. However once we get x=a,b
I used this info in st 1

a+b = -1

Since a=x and b=x

X+x = -1
x= -1/2

I know what I’m doing is not right but theoretically I cannot seem to understand why this approach is incorrect

Thanks

Posted from my mobile device
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,379
Own Kudos:
778,194
 [1]
Given Kudos: 99,977
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: 105,379
Kudos: 778,194
 [1]
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 20 Aug 2019, 21:44
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
578vishnu
Bunuel
578vishnu
Hi All,

can somebody please help me understand this? Here, when we get (x-a)(x-b) = 0, then why aren't the roots x=a and x=b ?

Then statement 1 will be sufficient ?

Please help!

VeritasKarishma Bunuel egmat chetan2u

This is a value question, which means that for a statement(s) to be sufficient you need to get single numerical value of the intersect points.

Hi Bunuel

I agree. However once we get x=a,b
I used this info in st 1

a+b = -1

Since a=x and b=x

X+x = -1
x= -1/2

I know what I’m doing is not right but theoretically I cannot seem to understand why this approach is incorrect

Thanks

Posted from my mobile device
Show more

x-intercepts of a parabola y = (x + a)(x + b) are x = -a and x = -b. This does not mean that -a = -b. For example, The x-intercepts of y = (x + 1)(x + 2) are x = -1 and x = -2.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,989
 [2]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,989
 [2]
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 20 Aug 2019, 22:44
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
578vishnu
Bunuel
578vishnu
Hi All,

can somebody please help me understand this? Here, when we get (x-a)(x-b) = 0, then why aren't the roots x=a and x=b ?

Then statement 1 will be sufficient ?

Please help!

VeritasKarishma Bunuel egmat chetan2u

This is a value question, which means that for a statement(s) to be sufficient you need to get single numerical value of the intersect points.

Hi Bunuel

I agree. However once we get x=a,b
I used this info in st 1

a+b = -1

Since a=x and b=x

X+x = -1
x= -1/2

I know what I’m doing is not right but theoretically I cannot seem to understand why this approach is incorrect

Thanks

Posted from my mobile device
Show more

In addition to what Bunuel said above, think in this way too:

When we have (x + a)(x + b) = 0, x is a variable and a and b are some constant values.

We know that x must take one of two values: -a or -b
x must be either -a or -b so that the product becomes 0.

When you say a = -x, and b = -x, you are assuming that a and b are equal and their value is some constant -x (which is not the case)

e.g.
(x + 2)(x - 1) = 0 implies x is either -2 or 1. Can I say that -2 = x and 1 = x so (-2 + 1) = x + x = -1?
This gives us x = -1/2 but we know that x is either -2 or 1. Then this is not possible. Our assumption here is that a and b are equal which they needn't be.
avatar
Hanzel101
avatar
Joined: 15 Jul 2016
Last visit: 23 Feb 2020
Posts: 72
Own Kudos:
92
Given Kudos: 190
Location: India
Schools: Oxford "21 (A) Cambridge "21 (A$)
GMAT 1: 690 Q48 V36
Schools: Oxford "21 (A) Cambridge "21 (A$)
GMAT 1: 690 Q48 V36
Posts: 72
Kudos: 92
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 21 Aug 2019, 00:08
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thankyou Bunuel and VeritasKarishma

It's clear now :)
avatar
mahsanmalik
avatar
Joined: 18 May 2019
Last visit: 14 Oct 2019
Posts: 3
Posts: 3
Kudos: 0
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 06 Sep 2019, 07:23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I stopped when I had two equation involving A&B thinking A and B will have multiple answer and close E.

For DS do we always need to find complete solution bec looking at the equation one might interpret that this equation would have multiple ans hence information is not sufficient.

VeritasKarishma
udaymathapati
Attachment:
M-Q19.JPG

At what two points does the graph of y = (x+a)(x+b) intersect the x axis?

You don't need to worry what the equation represents. Just think, what does 'intersection with x axis' imply? It means the y co-ordinate is 0.

0 = (x+a)(x+b)
or x = -a or -b
Hence the graph must intersect the x axis at points (-a, 0) and (-b, 0). We need the values of a and b now.

Statement 1: a + b = -1
Two variables, only one equation. Not sufficient.

Statement 2: Graph intersects the y axis at (0, -6).
At y axis, x = 0. This means when x = 0, y co-ordinate is -6.
Put these values in y = (x+a)(x+b) to get -6 = ab.
Again, two variables, one equation. Not sufficient alone.

Using both statements, we have two variables and two different equations so we will be able to find the values of a and b. It doesn't matter which is 'a' and which is 'b'. We find that the two of them are -3 and 2. Since we need the points (-a, 0) and (-b, 0), the required points are (3, 0) and (-2, 0). Sufficient.

Answer (C).
Show more
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 19 Nov 2025
Posts: 5,794
Own Kudos:
5,509
 [1]
Given Kudos: 161
Location: India
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
My Reviews
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
Posts: 5,794
Kudos: 5,509
 [1]
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 16 Sep 2019, 07:36
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
misterJJ2u
In the xy-plane, at what two points does the graph of y = (x + a)(x + b) intersect the x-axis?

(1) a + b = -1
(2) The graph intersects the y-axis at (0, -6)
Show more

Asked: In the xy-plane, at what two points does the graph of y = (x + a)(x + b) intersect the x-axis?

At x = -a & at x = -b, the graph intersect x-axis.

(1) a + b = -1
Multiple values of a & b satisfy the condition.
NOT SUFFICIENT

(2) The graph intersects the y-axis at (0, -6)
At x = 0 ; y=ab = -6
ab = -6
Multiple values of a & b satisfy the condition
NOT SUFFICIENT

(1) + (2)
(1) a + b = -1
(2) The graph intersects the y-axis at (0, -6)
ab = -6
b = -a -1
a(a+1) =6
a = 2 or -3
(a, b) = {(2,-3),(-3,2)}
The graph intersects x-axis at x=3 & x=-2
SUFFICIENT

IMO C
avatar
nishantkurup
avatar
Joined: 16 May 2020
Last visit: 28 Jul 2020
Posts: 6
Own Kudos:
4
Given Kudos: 23
Posts: 6
Kudos: 4
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 26 May 2020, 12:23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The general equation of a parabola is y = Ax^2 + Bx + C
The roots of this quadratic equation will give you the points at which the parabola intersects the x-axis.

When you simplify the question stem, you are left with y = x^2 + (a+b)x + ab. So in this case;
A = 1
B = (a+b)
C = (ab)

Hence if you need to find the roots, you need the values of all 3; A, B and C.
S1 gives you the value of B (which is a+b). Not sufficient since you dont have the value of C (which is ab)
S2 gives you the value of C (which is ab). Not sufficient since you dont have the value of B (which is a+b).

When you combine the 2 statements, you have a full parabola equation in the form of y = x^2 - x - 6
Hence your answer is C.

Note: Since its DS, you dont have to solve all the way, but for reference the solution would be (x-3)(x+2); so x= 3 and -2.
avatar
Usynlige
avatar
Joined: 25 Jul 2020
Last visit: 28 Jul 2020
Posts: 1
Posts: 1
Kudos: 0
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 28 Jul 2020, 05:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
In the xy-plane, at what two points does the graph of y = (x + a)(x + b) intersect the x-axis?

(1) a + b = -1
(2) The graph intersects the y-axis at (0, -6)

X-intercepts of the function \(f(x)\) or in our case the function (graph) \(y=(x+a)(x+b)\) is the value(s) of \(x\) for \(y=0\). So basically the question asks to find the roots of quadratic equation \((x+a)(x+b)=0\).

\((x+a)(x+b)=0\) --> \(x^2+bx+ax+ab=0\) --> \(x^2+(a+b)x+ab=0\).

Statement (1) gives the value of \(a+b\), but we don't know the value of \(ab\) to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of \(y\) when \(x=0\) --> \(y=(x+a)(x+b)=(0+a)(0+b)=ab=-6\). We know the value of \(ab\) but we don't know the value of \(a+b\) to solve the equation.

Together we know the values of both \(a+b\) and \(ab\), hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.


Answer: C.


Hope it's clear.
Show more


Clearly this kind of question is going to be a precious time waster.

Is there a short cut for this type other than just random guessing?
avatar
dortinator1234923
avatar
Joined: 08 Sep 2020
Last visit: 21 Jan 2021
Posts: 21
Own Kudos:
6
Given Kudos: 23
Location: Netherlands
Schools:  (S)
GMAT 1: 580 Q45 V25
GPA: 3.4
Schools:  (S)
GMAT 1: 580 Q45 V25
Posts: 21
Kudos: 6
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 10 Nov 2020, 11:01
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1) a +b = -1
-> a=-2

Insufficient

2) Intersects at (0,-6)
-> a * b = -6
Insufficient

1)+2) = -2*b+-6
b=3

We have value for a = -2 and b=3
User avatar
MHIKER
User avatar
Joined: 14 Jul 2010
Last visit: 24 May 2021
Posts: 942
Own Kudos:
5,645
Given Kudos: 690
Status:No dream is too large, no dreamer is too small
Concentration: Accounting
Posts: 942
Kudos: 5,645
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) Updated on: 13 Dec 2020, 12:21
Originally posted by MHIKER on 11 Nov 2020, 06:43.
Last edited by MHIKER on 13 Dec 2020, 12:21, edited 1 time in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
misterJJ2u
In the xy-plane, at what two points does the graph of y = (x + a)(x + b) intersect the x-axis?

(1) a + b = -1
(2) The graph intersects the y-axis at (0, -6)
Show more


This is a quadric equation. If we manipulate it then we get:
\(y=x^2+bx+ax+ab\)

\(y=x^2+x(a+b)+ab\)

To plot the graph we need the value of \((a+b)\) and \(ab\)

(1) Gives the value of (a+b) only INSUFFICIENT

(2) Gives the value of ab only
Putting the values in the equation
\(-6=0^2+0(a+b)+ab\)

\(ab=-6\)

INSUFFICIENT.

Using information from both the option we get values of a+b and ab. SUFFICIENT.

ANS. C
User avatar
Basshead
User avatar
Joined: 09 Jan 2020
Last visit: 07 Feb 2024
Posts: 925
Own Kudos:
302
 [1]
Given Kudos: 432
Location: United States
Posts: 925
Kudos: 302
 [1]
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 15 Nov 2020, 16:14
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
misterJJ2u
In the xy-plane, at what two points does the graph of y = (x + a)(x + b) intersect the x-axis?

(1) a + b = -1
(2) The graph intersects the y-axis at (0, -6)
Show more

\(y = x^2 + bx + ax + ab \)
\(y = x^2+ x(a+b) + ab\)

We need the value of a + b and ab.

(1) We don't know the value of ab; INSUFFICIENT.

(2) \(-6 = x^2+ 0(a+b) + ab\)
\(-6 = ab\)
We don't know the value of a+b; INSUFFICIENT.

(1&2) Together, we know ab and a+b; SUFFICIENT.

Answer is C.
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 18 Nov 2025
Posts: 21,712
Own Kudos:
26,995
 [1]
Given Kudos: 300
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,712
Kudos: 26,995
 [1]
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 13 Dec 2020, 05:35
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
misterJJ2u
In the xy-plane, at what two points does the graph of y = (x + a)(x + b) intersect the x-axis?

(1) a + b = -1
(2) The graph intersects the y-axis at (0, -6)
Show more
Solution:

The two points where the graph of y = (x + a)(x + b) intersects the x-axis are the two x-intercepts of the graph. To determine the x-intercepts, we solve for x when y = 0:

0 = (x + a)(x + b)

x = -a or x = -b

We see that if we can determine the values of a and b, then we can determine the two x-intercepts, namely, -a and -b.

Statement One Alone:

a + b = -1

Since there are many pairs of values that can add to -1, we can’t determine the exact values of a and b. Statement one alone is not sufficient.

Statement Two Alone:

The graph intersects the y-axis at (0, -6).

This means when x = 0, y = -6. That is:

-6 = (0 + a)(0 + b)

-6 = ab

Like statement one, there are many pairs of values that can multiply to -6, so we can’t determine the specific values of a and b. Statement two alone is not sufficient.

Statements One and Two Together:

From the two statements, we see that a + b = -1 and ab = -6. If we let b = -6/a and substitute it into the first equation, we have:

a + (-6/a) = -1

a^2 - 6 = -a

a^2 + a - 6 = 0

(a + 3)(a - 2) = 0

a = -3 or a = 2

Although it seems there are two distinct values for a instead of one unique value, let’s look at the corresponding values of b also.

Case 1: If a = -3, b = -6/(-3) = 2.

Case 2: If a = 2, b = -6/2 = -3.

Recall that we are looking for the two x-intercepts, which have the values of -a and -b. In the first case, we have the two x-intercepts as 3 and -2 and in the second, we also have the same two x-intercepts, -2 and 3. Therefore, both statements together are sufficient to answer the question.

Answer: C
User avatar
CrackverbalGMAT
User avatar
Major Poster
Joined: 03 Oct 2013
Last visit: 19 Nov 2025
Posts: 4,844
Own Kudos:
8,945
Given Kudos: 225
Affiliations: CrackVerbal
Location: India
Expert
Expert reply
Posts: 4,844
Kudos: 8,945
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 27 May 2021, 03:15
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Solution:

The equation in question here is (x+a)(x+b)=y

=> x^2 + bx+ ax + ab = y

=> x^2 + x(a+b) + ab = y

This graph would intersect the x-axis at points where x=0. At x=0 we can see y=ab.

St(1)- a + b = -1

There is no information about ab. Hence (Insufficient)

St(2)-The graph intersects the y-axis at (0, -6)

Since at x=0 y is -6,we can substitute the same in y=a b = -6 and have a b=-6

However, with st(2)alone, we have no information of (a+b) . Hence Insufficient

Combining both,

We know (a+b) and (ab).
Sufficient (option c)

Devmitra Sen
GMAT SME
User avatar
bv8562
User avatar
Joined: 01 Dec 2020
Last visit: 01 Oct 2025
Posts: 423
Own Kudos:
490
Given Kudos: 360
GMAT 1: 680 Q48 V35
GMAT 1: 680 Q48 V35
Posts: 423
Kudos: 490
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 07 Jul 2021, 05:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
If the question had asked "AT HOW MANY POINTS DOES THE GRAPH OF y = (x + a)(x + b) INTERSECT X-AXIS?" then "D" would have been the answer. Am I right?
User avatar
jpterrazas
User avatar
Joined: 10 Jan 2022
Last visit: 27 Jan 2023
Posts: 1
Given Kudos: 67
Posts: 1
Kudos: 0
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 21 Aug 2022, 16:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The graph \(y = (x + a)(x + b)\) intersects the x-axis at the points:

\(P1 = (-a, 0)\)
\(P2 = (-b,0)\)

So finding "the x-axis intersect" becomes "find \(a\) and \(b\)"

(1) \(a+b = -1\)
One equation for two variables \(a\) and \(b\) (Not Sufficient)

(2) If we substitute point \((0, -6)\) in the original graph we get
\(-6 = (0 + a)(0 + b)\)
\(-6 = ab\)
One equation for two variables \(a\) and \(b\) (Not Sufficient)

(1)+(2) We have now two variables, two equations
\(a+b = -1\)
\(-6 = ab\)

You can easily solve this (don't need to), but the answers would be \(a=2\) and \(b=-3\)

Therefore, the "the x-axis intersect" are points:

\(P1 = (-2, 0)\)
\(P2 = (3,0)\)

This is sufficient, so answer would be (C)
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,583
Own Kudos:
1,079
Posts: 38,583
Kudos: 1,079
In the xy-plane, at what two points does the graph of y=(x+a)(x+b) 18 Sep 2025, 02:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
   1   2 
Moderators:
Bunuel
Math Expert
105379 posts
kingbucky
496 posts