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

 It is currently 21 Aug 2019, 05:10 ### 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. ### Request Expert Reply # Each coefficient in the equation ax2 + bx + c = 0 is determined by thr

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

### Hide Tags

Director  P
Joined: 19 Oct 2018
Posts: 774
Location: India
Each coefficient in the equation ax2 + bx + c = 0 is determined by thr  [#permalink]

### Show Tags

2
8 00:00

Difficulty:   75% (hard)

Question Stats: 42% (01:57) correct 58% (02:36) wrong based on 38 sessions

### HideShow timer Statistics

Each coefficient in the equation $$ax^2 + bx + c = 0$$ is determined by throwing an ordinary die. Find the probability that the equation will have equal roots.

A. 1/72
B. 1/54
C. 5/216
D. 1/36
E. 1/27
Senior Manager  P
Joined: 16 Jan 2019
Posts: 413
Location: India
Concentration: General Management
WE: Sales (Other)
Each coefficient in the equation ax2 + bx + c = 0 is determined by thr  [#permalink]

### Show Tags

2
Each of the coefficients must be an integer greater than 0 and less than 7

When the equation can be in the form $$(mx+n)^2$$ we get equal roots

We have

$$(x+1)^2=x^2+2x+1$$ -----{1,2,1}

$$(x+2)^2=x^2+4x+4$$ -----{1,4,4}

$$(2x+1)^2=4x^2+4x+1$$ -----{4,4,1}

In addition we have cases where b=a+c {2,4,2} and {3,6,3}

Total number of possibilities for the coefficients is 6*6*6

Therefore the required probability = $$\frac{5}{6*6*6} = \frac{5}{216}$$

Answer is (C)

Originally posted by firas92 on 19 Jun 2019, 18:52.
Last edited by firas92 on 20 Jun 2019, 02:39, edited 1 time in total.
Intern  B
Joined: 11 Jun 2019
Posts: 23
Location: India
Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr  [#permalink]

### Show Tags

2
1
Hi, can you please explain why the coefficients can't be B=6,a=3 and C=3, for ax^2+bx+c

Posted from my mobile device
Senior Manager  P
Joined: 16 Jan 2019
Posts: 413
Location: India
Concentration: General Management
WE: Sales (Other)
Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr  [#permalink]

### Show Tags

1
omkartadsare wrote:
Hi, can you please explain why the coefficients can't be B=6,a=3 and C=3, for ax^2+bx+c

Posted from my mobile device

You are right! Thanks for pointing that out.. I fell into a classic trap!

{2,4,2} and {3,6,3} also work.

I suppose the answer should be 5/216 then!
Intern  B
Joined: 08 Oct 2018
Posts: 4
Location: Canada
Concentration: Technology, Entrepreneurship
GMAT 1: 720 Q50 V38 WE: Information Technology (Non-Profit and Government)
Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr  [#permalink]

### Show Tags

3
Dividing the equation by a, we get x^2 + (b/a)x + (c/a) = 0, which must also satisfy (x+√ (c/a))^2 = 0 in order for the equation to have equal roots.

Expanding the 2nd equation gets x^2 + 2√ (c/a)x + (c/a) = 0, and comparing its 2nd term to the first equation's 2nd term, we get b/a = 2√ (c/a).
Since a, b and c must be integers from 1 to 6, we can square and manipulate the equation to get
(b/a)^2 = 4(c/a)
b^2 = 4ac
b^2 - 4ac = 0

b^2 - 4ac is the discriminant in the quadratic formula, and although GMAT doesn't require the knowledge of the quadratic formula, it certainly makes sense for it to equal 0 here in order to get equal roots.

Now, to calculate the probability of this equation holding, test with real values to see how many work.
b=2 -> a=1, c=1
b=4 -> a=1, c=4
b=4 -> a=4, c=1
b=4 -> a=2, c=2
b=6 -> a=3, c=3

So there are 5 combinations that work out of a total combination of 6*6*6 = 216 rolls.

Therefore the probability needed is 5/216. The answer is C.
Senior Manager  P
Joined: 15 Feb 2018
Posts: 299
Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr  [#permalink]

### Show Tags

Is there a way to solve this without having to test real values? I feel it would take me 62 minutes to just test them all...
I didn't connect 'equal roots' with b^2-4ac, great explanation
Intern  B
Joined: 08 Oct 2018
Posts: 4
Location: Canada
Concentration: Technology, Entrepreneurship
GMAT 1: 720 Q50 V38 WE: Information Technology (Non-Profit and Government)
Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr  [#permalink]

### Show Tags

1
It shouldn't take long to test with real values because the values can be derived.

4ac is even so b^2 has to be even, hence b has to be even.
b=2:
4=4ac
1=ac

b=4:
16=4ac
4=ac

b=6:
36=4ac
9=ac

It's just a matter of solving these 3 equations. Hope this helps.

Posted from my mobile device Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr   [#permalink] 09 Aug 2019, 18:20
Display posts from previous: Sort by

# Each coefficient in the equation ax2 + bx + c = 0 is determined by thr

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

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

#### MBA Resources  