GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Oct 2019, 07:55 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

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.  Does the curve h(x)=y=ax^2+b intersects the x-axis?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Manager  G
Joined: 15 Dec 2015
Posts: 114
GMAT 1: 680 Q49 V34 GPA: 4
WE: Information Technology (Computer Software)
Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

Show Tags

1
8 00:00

Difficulty:   75% (hard)

Question Stats: 63% (01:46) correct 37% (01:45) wrong based on 91 sessions

HideShow timer Statistics

Does the curve h(x)=y=ax^2+b intersects the x-axis?

Statement 1: h(4)>0
Statement 2: a>0

Originally posted by DHAR on 30 Jul 2017, 21:54.
Last edited by Bunuel on 17 Feb 2019, 02:17, edited 1 time in total.
Edited the OA.
Manager  S
Joined: 03 Jan 2016
Posts: 57
Location: India
WE: Engineering (Energy and Utilities)
Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

Show Tags

2
Hi,

Question :

Does the curve h(x)=y=ax^2+b intersects the x-axis?

Basically its a parabola equation.

For any parabola, ax2 + bx + C = 0, it will intersect the x axis only if discriminant b2-4ac = 0.

it wont intersect x axis if b2- 4ac<0;

Coming to the question:

y = ax2 + b
dicriminant = -4ab ;
now our aim is to check whether -4ab (discriminant in the given equation) is less than, equal or greater than 0.
1) h(4)>0 = > 16a+b>0 = >from this we cannot say about a & b
we dont know whether a and b is + ve or -ve.

So insuff

2) a>0
so we now have a(>0), but we dont know the value of b.

So insuff.

Combining 1 & 2
16a + b > 0
& a>0

therefore we can infer that b >0

so our discriminant -4ab<0

so it dont have any x intercepts.

Suff

C

Regards
Raju
Current Student B
Joined: 11 May 2015
Posts: 34
Location: United States
Concentration: Strategy, Operations
GPA: 3.44
Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

Show Tags

gvvsnraju@1 wrote:
Hi,

Combining 1 & 2
16a + b > 0
& a>0

therefore we can infer that b >0

so our discriminant -4ab<0

what if a = 1 and b = -2
Intern  B
Joined: 26 Oct 2010
Posts: 14
Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

Show Tags

I think the answer should be E.

the question asks us to check if -4ab is negative.

(1): 16a + b > 0

16(1) + 2 > 0; 16(1) - 2 > 0; 16(-1) + 17 > 0

SO, Not sufficient

(2) a>0. Nothing about b. So, -4ab can be positive or negative. Not sufficient.

(1) + (2): Still not sufficient since b can be + or -.

Ans: E.

Bunuel, please let us know if this approach is correct.
Manager  G
Joined: 23 Jan 2018
Posts: 221
Location: India
WE: Information Technology (Computer Software)
Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

Show Tags

chetan2u, Bunuel

Dear experts: Please check if C is the correct answer. I think E should be the answer.

Regards,
Arup
Intern  B
Joined: 13 Jun 2018
Posts: 37
GMAT 1: 700 Q49 V36 Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

Show Tags

Statement 1: 16a + b > 0

Both a and b are unknown and thus, insufficient

Statement 2: a > 0

This tells us ax^2 is always positive, however b could be negative. Insufficient

Both 1 and 2 together:

16a + b is positive and that 16a is positive.

B could be positive, no intersection or B could be a large negative number which could give us 1 or 2 intersections. Therefore insufficient.

E
Director  G
Joined: 09 Mar 2018
Posts: 994
Location: India
Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

Show Tags

GMATaspirant641 wrote:
I think the answer should be E.

the question asks us to check if -4ab is negative.

(1): 16a + b > 0

16(1) + 2 > 0; 16(1) - 2 > 0; 16(-1) + 17 > 0

SO, Not sufficient

(2) a>0. Nothing about b. So, -4ab can be positive or negative. Not sufficient.

(1) + (2): Still not sufficient since b can be + or -.

Ans: E.

Bunuel, please let us know if this approach is correct.

Hey GMATaspirant641

After combining when we use a>0 in 16a + b > 0, both cases will be valid, b can be <0 or b can be > 0

*Editing it*

Now if -4ab < 0

This means that determinant can be greater than 0 or less than 0

Giving answer as ,rightly mentioned in previous posts, E
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Originally posted by KanishkM on 16 Feb 2019, 12:17.
Last edited by KanishkM on 16 Feb 2019, 19:22, edited 1 time in total.
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1806
Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

Show Tags

2
Two posts above suggest that these inequalities:

16a + b > 0
a > 0

imply that b > 0. That is not the case, as you can see by plugging in numbers: we might have a = 1 and b = -1, for example.

If our parabola is y = ax^2 + b, then when a is positive, the parabola (a U shape) opens upwards, and if a is negative, the parabola opens downwards (an upside-down U shape). The value of b is the y-intercept of the parabola. Using both statements here, we know the parabola opens upwards, because a > 0, and we know that when x=4, the parabola is above the x-axis. But that still leaves two possibilities: maybe the parabola has a y-intercept that is above the origin (b > 0), and lies entirely above the x-axis (e.g. in the parabola y = x^2 + 1) or maybe it starts at or below the x-axis (b < 0), crosses the x-axis in one or two places, and becomes positive by the time x = 4 (e.g. in the parabola y = x^2 - 1, which clearly has two x-intercepts at 1 and -1). So the answer is E.

If the source is claiming the answer is C here, then I'd suggest working with more reliable study materials.
_________________
GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?   [#permalink] 16 Feb 2019, 15:52
Display posts from previous: Sort by

Does the curve h(x)=y=ax^2+b intersects the x-axis?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  