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Senior Manager  Joined: 29 Nov 2006
Posts: 284
Location: Orange County, CA
In the xy-plane, at what two points does the graph of  [#permalink]

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148 00:00

Difficulty:   35% (medium)

Question Stats: 71% (01:50) correct 29% (01:57) wrong based on 1948 sessions

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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)

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-xy-plane-at-what-points-does-the-graph-of-y-88398.html

Originally posted by misterJJ2u on 21 Jan 2007, 00:36.
Last edited by Bunuel on 01 Jun 2013, 03:28, edited 2 times in total.
Edited the question and added the OA
##### Most Helpful Expert Reply
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: graph question - gmat prep  [#permalink]

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95
89
In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis?
(1) a+b=-1
(2) The graph intersects 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.
_________________

Originally posted by Bunuel on 22 Oct 2009, 13:28.
Last edited by Bunuel on 22 Oct 2009, 13:47, edited 1 time in total.
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SVP  Joined: 01 May 2006
Posts: 1740

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15
12
For me (C) y= (x+a)(x+b) intersect the x-axis is equivalent to search the solutions of :
(x+a)(x+b) =0
<=> x^2 + (a+b)*x + a*b = 0 (1)

From 1
a+b = -1
implies:

(1) <=> x^2 -x + a*(-a-1) = 0

we need to know a.

INSUFF.

From 2
(0, -6) is the Y interceptor. That implies:

a*b = -6.

we need another equation with a & b, in other words, the values of a and b.

INSUFF.

Both (1) and (2)
a+b = -1
a*b = -6

These equations give us (a,b), what we need to find the roots.

SUFF.
##### General Discussion
Senior Manager  Joined: 06 Apr 2008
Posts: 361
Re: DS: Coordinate Geometry (intersection)  [#permalink]

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2
lordw wrote:
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).

Guys I felt lost in this problem with all the letters. Any idea of how to put this in a more "visible" way? For instance, may be picking numbers? Many tks. Lw.

Graph intersects x-axis when y=0 , therefore two points where graph intersects is x=-a, x=-b

Statement 1) is insuff

Statement 2) just tells one point

Combining both , points are 6 and -5

Answer is C)
Senior Manager  Joined: 19 Mar 2008
Posts: 332
Re: DS: Coordinate Geometry (intersection)  [#permalink]

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3
1
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).

for point intersect the x axis, we need to find x when y = 0
so 0 = (x+a) (x+b)
x = -a or -b
So, we need to find a and b

(1) +(2) is sufficient to find two unknown

(C)
Manager  Joined: 27 Sep 2008
Posts: 73
Re: xy plane  [#permalink]

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3
The answer is (E)

y=(x+a)(x+b)

y=x^2+ax+ab+ab

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

statement 1

a+b = -1

y=x^2-x+ab

0=x^2-x+ab

insufficient

statement 2

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

ab=-6

0=x^2+x(a+b)-6

insufficient

both statements

0=x^2-x-6

6=x(x-1)

x can be -2 or 3

insufficient Senior Manager  Joined: 21 Apr 2008
Posts: 256
Location: Motortown
Re: xy plane  [#permalink]

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I agree with fresinha12, I get 3 and -2

in the question, it asks for points, so I guess it is okay to say that the answer is C
VP  Joined: 17 Jun 2008
Posts: 1162
Re: xy plane  [#permalink]

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1
redbeanaddict wrote:
In the xy plane, at what 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)

the graph intersects x axis hen y=0 => x=-a and x=-b are the points

to get the points we must know a and b but how to find

1) does not help
2)does not help => gives ab=-6

combine 1 and 2 we get a and b hence is SUFFI to answr
IMO C
_________________
cheers
Its Now Or Never
Manager  Joined: 22 Jul 2009
Posts: 161
Re: GMAT PREP_DS_XY plane  [#permalink]

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The answer provided above is incorrect.

The question does not ask for the values of a and b.

It asks for the values of x when y=0. In other words, it asks for the roots of the quadratic.

Statement 2 is sufficient, as it says that ab=-6.

Thus: x^2-6x-6=0 -> Roots are -0.87 and 6.87.

Answer is B.
Manager  Joined: 29 Jul 2009
Posts: 157
graph question - gmat prep  [#permalink]

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1
Hi,

so here's what I did:

y= x^2+ xb+xa+ab

I replace y with -6 and x with 0

-6=0+0+ab
so ab= -6

I replace ab with -6 and y with 0 in the first equation

0= x^2+ xb+xa+(-6)
6=x(x+b+a)

from statement A, we know that a+b=1
so 6 = x(x+1)
x^2+x-6 = 0

(x-3)(x+2)
x=3 or x = -2

I then got stuck so what is the value of x?
Attachments Q19-DS.jpg [ 49.7 KiB | Viewed 124589 times ]

Manager  Joined: 29 Jul 2009
Posts: 157
Re: graph question - gmat prep  [#permalink]

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1
Bunuel wrote:
In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis?
(1) a+b=-1
(2) The graph intersects y-axis at (0,-6)

Q is in what two points graph intersect the x-axis. Which means we should be able to find two points: (x1,0) and (x2,0), so basiacally the roots of equation: (x+a)(x+b)=0.

y=(x+a)(x+b)=x^2+x(a+b)+ab. --> We must now the values of (a+b) and ab.

(1) a+b=-1 --> not sufficient.
(2) x=0, y=-6 --> -6=ab not sufficient

(1)+(2) a+b=-1, ab=-6 --> y=x^2+x(a+b)+ab=x^2-x-6 --> x intersection means y=0 --> 0=x^2-x-6 --> (x+2)(x-3)=0 -
-> x1=-2, x2=3. Sufficient

Answer C.

OMG!!

I didnt realize that I need to find two points!!
thats why when i ended up with -2 and 3, i got confused because i was left with two values!!!

how did you find ab=-6?
(2) x=0, y=-6 --> -6=ab not sufficient

did you use the formula
y=mx+b?

thanks,
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: graph question - gmat prep  [#permalink]

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2
(2) The graph intersects y-axis at (0,-6) --> x=0, y=-6 --> y=x^2+x(a+b)+ab --> -6=0^2+0(a+b)+ab --> ab=-6.
_________________
Manager  Joined: 05 Jul 2009
Posts: 152
Re: graph question - gmat prep  [#permalink]

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Great Bunuel! I have been seeing your post and I have to say I am quite amazed at your math skills! Have you taken the GMAT yet? I would very much like to know your score.
Director  Joined: 01 Apr 2008
Posts: 739
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Re: graph question - gmat prep  [#permalink]

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Yup Bunuel is awesome as always Here, we need to find the solution for (x+a)(x+b) = 0.
Solutions are x=-a, x=-b.
Thus we need to find unique values for a and b to answer the question.

1) we get a relation between a and b but not sufficient to derive unique values of a and b. insuff.

2) we get a point on the graph, so substituting values of x and y in the graph we get ab = -6. Again, this is insufficient to find unique values of a and b. insuff.

Combining, we have, a+b=-1 and ab= -6, two variables, two equations. So answer is C. The points will be (-a,0), (-b,0).
Manager  Joined: 22 Sep 2009
Posts: 183
Location: Tokyo, Japan
Re: GMAT PREP_DS_XY plane  [#permalink]

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powerka wrote:
The answer provided above is incorrect.

The question does not ask for the values of a and b.

It asks for the values of x when y=0. In other words, it asks for the roots of the quadratic.

Statement 2 is sufficient, as it says that ab=-6.

Thus: x^2-6x-6=0 -> Roots are -0.87 and 6.87.

Answer is B.

Fig is correct. You made a mistake on the equation, it is x^2-(a+b)x+ab.

Essentially, you are looking for a and b itself, or equations that tell you a+b and ab.

Hence, C
Intern  Joined: 04 Dec 2009
Posts: 17
Re: graph question - gmat prep  [#permalink]

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Bunuel wrote:
In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis?
(1) a+b=-1
(2) The graph intersects 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.

I get your explanation.But using a+b=-1 and ab=-6 and solving it I get (a-b)^2=25 .which means a-b=+5 or -5.So since I cant be sure of the value of a-b,I selected option e as the answer.Can you please help on this.
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: graph question - gmat prep  [#permalink]

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8
13
gautamsubrahmanyam wrote:
Bunuel wrote:
In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis?
(1) a+b=-1
(2) The graph intersects 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.

I get your explanation.But using a+b=-1 and ab=-6 and solving it I get (a-b)^2=25 .which means a-b=+5 or -5.So since I cant be sure of the value of a-b,I selected option e as the answer. Can you please help on this.

The question asks: at what two points the graph of $$y=(x+a)(x+b)$$ intersect the x-axis?

So we should find two points: (x1, 0) and (x2, 0), (points of intersection of the given graph with X-axis). Basically the question asks to find the roots of quadratic equation $$(x+a)(x+b)=0$$ --> $$x^2+(a+b)x+ab=0$$.

When we combine statement we have: $$a+b=-1$$ and $$ab=-6$$, so you should solve quadratic equation $$x^2-x-6=0$$ --> $$x_1=-2$$ and $$x_2=3$$ --> points of intersection are (-2, 0) and (3, 0). Check on the diagram:
Attachment: MSP272919df1bed26600c1g0000673adafe5fce1fc2.gif [ 3.61 KiB | Viewed 123077 times ]
Hope it's clear.
_________________
Manager  Joined: 19 Dec 2010
Posts: 95
Re: xy plane  [#permalink]

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This problem is easy when you don't start by rapidly trying to solve it.
Think about it before you put pen to paper...
"what two points does the graph intersect the x axis..."
This is a simple problem asking for 2 values of x...
statement 1 gives us a+b=-1
statement 2 gives us ab

Start with the stem...multiply x+a and x+b, it will become very clear.
Y = X^2 + X (a+b) + ab
statement 1 gives you a+b, we need ab
statement 2 says graph intersects y at (0,-6). Plug this into the original statement, you will get ab.
Solved = C! Re: xy plane   [#permalink] 17 Mar 2011, 22:27
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