NEW HERE? LEARN MORE
closeclose
It is currently 19 Apr 2024, 09:14
Register
GMAT Focus Tests
Decision Tracker
My Rewards
New posts
New comers' posts

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

M28-08

12 posts Go to First Unread Post
LOCKED TOPIC
Question banks

Question Banks

Downloads
My Bookmarks
Reviews
SORT BY:
Date
Kudos
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92758
CAT Tests Forum Quiz Mode
M28-08 [#permalink] New post  16 Sep 2014, 01:28
7
Bookmarks
Expert Reply
00:00
A
B
C
D
E

Difficulty:

55% (hard)

Question Stats:

62% (01:59) correct 38% (01:54) wrong based on 127 sessions
Hide Show timer Statistics
The equation \(x^2 + ax - b = 0\) has equal roots, and one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3. What is the value of \(b\)?

A. \(-64\)
B. \(-16\)
C. \(-15\)
D. \(-\frac{1}{16}\)
E. \(-\frac{1}{64}\)
ShowHide Answer
Official Answer
Official Answer and Stats are available only to registered users. Register/Login.

_________________
New to the GMAT CLUB Forum?

My Signature Questions' Collection:


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
Signature Read More
Most Helpful Reply
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92758
CAT Tests Forum Quiz Mode
Re M28-08 [#permalink] New post  16 Sep 2014, 01:28
4
Kudos
1
Bookmarks
Expert Reply
Official Solution:

The equation \(x^2 + ax - b = 0\) has equal roots, and one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3. What is the value of \(b\)?

A. \(-64\)
B. \(-16\)
C. \(-15\)
D. \(-\frac{1}{16}\)
E. \(-\frac{1}{64}\)


Since one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3, then substituting we'll get: \(3^3+3a+15=0\). Solving for \(a\) gives \(a=-8\).

Substitute \(a=-8\) in the first equation: \(x^2-8x-b=0\).

Now, we know that it has equal roots thus its discriminant must equal to zero: \(d=(-8)^2+4b=0\). Solving for \(b\) gives \(b=-16\).


Answer: B
_________________
New to the GMAT CLUB Forum?

My Signature Questions' Collection:


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
Signature Read More
General Discussion
User avatar
JackSparr0w
Manager
Manager
Joined: 08 Feb 2014
Posts: 168
Location: United States
Concentration: Finance
Schools: Cnsrtm-Stern '17 (M$) Cnsrtm-Darden '17 (WL) Cnsrtm-Johnson '17 (A) Cnsrtm-Tuck '17 (WL)
GMAT 1: 650 Q39 V41
WE:Analyst (Commercial Banking)
Re: M28-08 [#permalink] New post  04 Nov 2014, 12:38
I thought the Discriminant was (b^2)-4ac.

This would yeild ((-8)^2)-4(1)(b) --> 64-4x.

So: 64-4x=0 -->64=4x -->x=16.

Can someone explain why this isn't the case? In Bunuel's solution we have addition, instead of subtraction in the discriminant.

Thanks
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92758
CAT Tests Forum Quiz Mode
Re: M28-08 [#permalink] New post  05 Nov 2014, 05:28
1
Kudos
1
Bookmarks
Expert Reply
JackSparr0w wrote:
I thought the Discriminant was (b^2)-4ac.

This would yeild ((-8)^2)-4(1)(b) --> 64-4x.

So: 64-4x=0 -->64=4x -->x=16.

Can someone explain why this isn't the case? In Bunuel's solution we have addition, instead of subtraction in the discriminant.

Thanks


Yes, the discriminant of \(ax^2 + bx + c = 0\) is \(b^2 - 4ac\).

Thus the discriminant of \(x^2 - 8x - b = 0\) is : \((-8)^2 - 4(-b) = (-8)^2 + 4b\).

Hope it's clear.
_________________
New to the GMAT CLUB Forum?

My Signature Questions' Collection:


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
Signature Read More
User avatar
JackSparr0w
Manager
Manager
Joined: 08 Feb 2014
Posts: 168
Location: United States
Concentration: Finance
Schools: Cnsrtm-Stern '17 (M$) Cnsrtm-Darden '17 (WL) Cnsrtm-Johnson '17 (A) Cnsrtm-Tuck '17 (WL)
GMAT 1: 650 Q39 V41
WE:Analyst (Commercial Banking)
Re: M28-08 [#permalink] New post  05 Nov 2014, 07:58
Got it, thanks!
avatar
filipTGIM
Intern
Intern
Joined: 21 May 2013
Posts: 5
Re: M28-08 [#permalink] New post  25 Dec 2014, 11:22
Isn't it just more simple to acknowledge that x^2-8x-b=0 and then, knowing the roots are same, use reverse foil method find: 4+4=8, 4*4=16. ??

Knowing that Bunuel always uses the simplest method, please explain the importance of knowing the discriminant? I thought with reverse foil the discriminant was not important to remember.

Thanks !
User avatar
propcandy
Intern
Intern
Joined: 29 Jun 2014
Posts: 6
GMAT ToolKit User
Re: M28-08 [#permalink] New post  21 Mar 2015, 10:50
1
Kudos
This question was duplicated from Q D01-41. I got this question twice from a quiz I was doing.
User avatar
B
HimJ
Intern
Intern
Joined: 15 Dec 2016
Posts: 7
Location: India
GMAT 1: 600 Q44 V28
GMAT 2: 700 Q49 V37
GPA: 3.9
WE:Marketing (Real Estate)
Reviews Badge
Re: M28-08 [#permalink] New post  18 Oct 2018, 21:44
1
Kudos
Alternative method:


We know one of the roots of x^2 +ax + 15 = 0, so let x=3
a=-8 . . . . . by substituting x=3 in x^2 +ax + 15 = 0

Standard form of quadratic equation is ax^2+bx+c=0, let its root be p and q. so, (x-p)(x-q)=0
Solving (x-p)(x-q)=0, we get x^2 -(p+q)+pq=0

By comparing x^2 -(p+q)+pq=0 with x^2 +ax - b = 0 that is x^2 -8x - b=0 . . . . .as a=-8
we can deduce that p+q=8 and pq=-b
It is given that the roots are equal so both the roots p and q must be 4 as p+q=8
and pq=16

Substituting above values in pq=-b we get that -b=16 so b=-16


If you learned something new, hit kudos button as a gratitude :)
avatar
B
apurvgodbole
Intern
Intern
Joined: 16 Apr 2017
Status:Attempting to cross the 750 barrier
Posts: 5
GMAT 1: 530 Q39 V24
Re: M28-08 [#permalink] New post  18 Oct 2018, 23:23
We know one of the roots of x^2 +ax + 15 = 0 is 3.
Substitute x=3 in the equation.
We get x= -8 as the other root.
We know that sum of roots in equation ax^2 +bx + c = 0, is -b/a and product of the roots is c/a.
So, in equation x^2 +ax -b = 0, a becomes -8 => equation becomes, x^2 -8x - b = 0
(Let root 1= root 2 = r)
r + r = -8 => r = -4
r*r = -b => b = -16

Ans B

I was comfortable using the sum and product of roots concept and could solve faster that way. Hence, I thought I might add this as a solution.

Cheers!
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92758
CAT Tests Forum Quiz Mode
Re: M28-08 [#permalink] New post  23 Feb 2023, 06:14
Expert Reply
Duplicate of D01-41. Unpublished.
_________________
New to the GMAT CLUB Forum?

My Signature Questions' Collection:


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
Signature Read More
User avatar
S
Dumsy_1711
Manager
Manager
Joined: 23 May 2023
Posts: 63
Location: India
Concentration: Real Estate, Sustainability
GPA: 3.7
WE:Other (Other)
M28-08 [#permalink] New post  23 Nov 2023, 08:43
Guys I have a doubt.

I tried solving by substituting the values for b. So for b = -16

\(x^2 -4x-4x -16 = 0\)

x (x-4) - 4(x+4) =0

But the equation cannot be factored this way. Can anyone tell me where I'm going wrong please?
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92758
CAT Tests Forum Quiz Mode
Re: M28-08 [#permalink] New post  23 Nov 2023, 08:54
Expert Reply
Dumsy_1711 wrote:
Guys I have a doubt.

I tried solving by substituting the values for b. So for b = -16

\(x^2 -4x-4x -16 = 0\)

x (x-4) - 4(x+4) =0

But the equation cannot be factored this way. Can anyone tell me where I'm going wrong please?


When you substitute n = -16 into x^2 - 8x - n = 0 you'd get x^2 - 8x - (-16) = 0, and finally, x^2 - 8x + 16 = 0, not x^2 - 8x - 16 = 0. And x^2 - 8x + 16 = 0 can be written as (x - 4)^2 = 0.

Hope it helps.

P.S. If still have any questions please continue discussion in this topic: https://gmatclub.com/forum/d01-183502.html

THIS TOPIC IS LOCKED.
_________________
New to the GMAT CLUB Forum?

My Signature Questions' Collection:


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
Signature Read More
GMAT Club Bot
gmatclubot
Re: M28-08 [#permalink]
23 Nov 2023, 08:54
LOCKED TOPIC
Downloads
My Bookmarks
Reviews
Moderator:
Bunuel
Math Expert
92758 posts

cron

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions