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

 It is currently 17 Feb 2019, 00:40

### 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

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Algebra Webinar

February 17, 2019

February 17, 2019

07:00 AM PST

09:00 AM PST

Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
• ### Valentine's day SALE is on! 25% off.

February 18, 2019

February 18, 2019

10:00 PM PST

11:00 PM PST

We don’t care what your relationship status this year - we love you just the way you are. AND we want you to crush the GMAT!

# M60-12

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6949
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

11 Jun 2018, 05:12
1
00:00

Difficulty:

45% (medium)

Question Stats:

69% (00:48) correct 31% (01:21) wrong based on 16 sessions

### HideShow timer Statistics

When $$a≠0$$, how many solutions does the equation $$a(x+b)^2+c=0$$ have?

1) $$bc=0$$

2) $$|b|+|c|=0$$

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only \$149 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Current Student
Joined: 19 Mar 2012
Posts: 4351
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)

### Show Tags

11 Jun 2018, 05:12
Official Solution:

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

There are three cases to consider:

Case 1: $$a > 0$$, $$c > 0$$ or $$a < 0$$, $$c < 0$$

The equation has no roots.

Case 2: $$c = 0$$.

The equation has only one root.

Case 3: $$a > 0$$, $$c < 0$$ or $$a < 0$$, $$c > 0$$

The equation has two roots.

Condition 1):

If $$bc = 0$$, then when

$$a = 1$$, $$b = 0$$, $$c = -1$$, the equation has two roots, and when

$$a = 1$$, $$b = 0$$, $$c = 0$$, the equation has one root.

As the question does not have a unique answer, condition 1) is not sufficient.

Condition 2)

$$|b| + |c| = 0$$ ⇔ $$b = c = 0$$.

Since $$c = 0$$, the equation has only one root.

Condition 2) is sufficient.

_________________
Intern
Joined: 16 Mar 2018
Posts: 6

### Show Tags

27 Jul 2018, 06:29
Not sure if I get the break down of the question. Can you please explain how you broke it down in a different way?
Intern
Joined: 12 Sep 2016
Posts: 12
GMAT 1: 640 Q44 V34
WE: Accounting (Consulting)

### Show Tags

03 Oct 2018, 04:24
I think this the explanation isn't clear enough, please elaborate. Can some one please explain how is statement 1 manipulated into the 3 cases. I am not able to follow how the three cases have been formed and why they have been formed?

Director
Joined: 08 Jun 2013
Posts: 552
Location: France
GMAT 1: 200 Q1 V1
GPA: 3.82
WE: Consulting (Other)

### Show Tags

03 Oct 2018, 08:39
2
souvik101990 wrote:
Official Solution:

Question prompt :

When a≠0, how many solutions does the equation $$a(x+b)^2+c=0$$ have?

1) bc=0

2) |b|+|c|=0

--> Solving $$a(x+b)^2+c=0$$ gives

$$(x+b)^2 = (-c/a)$$

Now lets discuss each of the 3 cases one by one :

There are three cases to consider:

Case 1: $$a > 0$$, $$c > 0$$ or $$a < 0$$, $$c < 0$$

The equation has no roots.

$$(x+b)^2 = (-c/a)$$ ---> LHS is positive number.

So we need RHS also positive.

$$a > 0$$, $$c > 0$$ ---> (-c/a) negative number.

$$a < 0$$, $$c < 0$$ ---> (-c/a) negative number.

So LHS can not be equal to RHS. No roots.

Case 2: $$c = 0$$.

The equation has only one root.

$$(x+b)^2 = (-c/a)$$

c= 0 then $$(x+b)^2 = 0$$

This gives one root x = -b.

Case 3: $$a > 0$$, $$c < 0$$ or $$a < 0$$, $$c > 0$$

The equation has two roots.

$$(x+b)^2 = (-c/a)$$ ---> LHS is positive number.

So we need RHS also positive.

$$a > 0$$, $$c < 0$$ ---> (-c/a) positive number. or

$$a < 0$$, $$c > 0$$ ---> (-c/a) positive number.

$$(x+b)^2 = (-c/a) = K$$

So roots will be $$x = -b +/- \sqrt{k}$$

Condition 1):

If $$bc = 0$$, then when

$$a = 1$$, $$b = 0$$, $$c = -1$$, the equation has two roots, and when

$$a = 1$$, $$b = 0$$, $$c = 0$$, the equation has one root.

As the question does not have a unique answer, condition 1) is not sufficient.

Condition 2)

$$|b| + |c| = 0$$ ⇔ $$b = c = 0$$.

Since $$c = 0$$, the equation has only one root.

Condition 2) is sufficient.

ridhi9jain

I have tried to breakdown the OE 3 cases...

Hope that helps!!
_________________

Everything will fall into place…

There is perfect timing for
everything and everyone.
Never doubt, But Work on
improving yourself,
Keep the faith and
It will all make sense.

Director
Joined: 08 Jun 2013
Posts: 552
Location: France
GMAT 1: 200 Q1 V1
GPA: 3.82
WE: Consulting (Other)

### Show Tags

03 Oct 2018, 08:43
ridhi9jain wrote:
I think this the explanation isn't clear enough, please elaborate. Can some one please explain how is statement 1 manipulated into the 3 cases. I am not able to follow how the three cases have been formed and why they have been formed?

souvik101990 wrote:
Official Solution:

Question prompt :

When a≠0, how many solutions does the equation $$a(x+b)^2+c=0$$ have?

1) bc=0

2) |b|+|c|=0

--> Solving $$a(x+b)^2+c=0$$ gives

$$(x+b)^2 = (-c/a)$$

Now lets discuss each of the 3 cases one by one :

There are three cases to consider:

Case 1: $$a > 0$$, $$c > 0$$ or $$a < 0$$, $$c < 0$$

The equation has no roots.

$$(x+b)^2 = (-c/a)$$ ---> LHS is positive number.

So we need RHS also positive.

$$a > 0$$, $$c > 0$$ ---> (-c/a) negative number.

$$a < 0$$, $$c < 0$$ ---> (-c/a) negative number.

So LHS can not be equal to RHS. No roots.

Case 2: $$c = 0$$.

The equation has only one root.

$$(x+b)^2 = (-c/a)$$

c= 0 then $$(x+b)^2 = 0$$

This gives one root x = -b.

Case 3: $$a > 0$$, $$c < 0$$ or $$a < 0$$, $$c > 0$$

The equation has two roots.

$$(x+b)^2 = (-c/a)$$ ---> LHS is positive number.

So we need RHS also positive.

$$a > 0$$, $$c < 0$$ ---> (-c/a) positive number. or

$$a < 0$$, $$c > 0$$ ---> (-c/a) positive number.

$$(x+b)^2 = (-c/a) = K$$

So roots will be $$x = -b +/- \sqrt{k}$$

Condition 1):

If $$bc = 0$$, then when

$$a = 1$$, $$b = 0$$, $$c = -1$$, the equation has two roots, and when

$$a = 1$$, $$b = 0$$, $$c = 0$$, the equation has one root.

As the question does not have a unique answer, condition 1) is not sufficient.

Condition 2)

$$|b| + |c| = 0$$ ⇔ $$b = c = 0$$.

Since $$c = 0$$, the equation has only one root.

Condition 2) is sufficient.

ridhi9jain

I have tried to breakdown the OE 3 cases...

Hope that helps!!
_________________

Everything will fall into place…

There is perfect timing for
everything and everyone.
Never doubt, But Work on
improving yourself,
Keep the faith and
It will all make sense.

Director
Joined: 08 Jun 2013
Posts: 552
Location: France
GMAT 1: 200 Q1 V1
GPA: 3.82
WE: Consulting (Other)

### Show Tags

03 Oct 2018, 08:51
Paugustin90 wrote:
Not sure if I get the break down of the question. Can you please explain how you broke it down in a different way?

Paugustin90

Question prompt :

When a≠0, how many solutions does the equation $$a(x+b)^2+c=0$$ have?

1) bc=0

2) |b|+|c|=0

St 1 : bc = 0 so b= 0 or c= 0 or both b and c = 0.

c=0 gives $$a(x+b)^2+0=0$$ ----> x = -b

Both b= 0 and c= 0 gives $$a(x+0)^2+0=0$$ -----> x =0

2 different answers. We can eliminate st 1. Option A and D are gone.

For sack of completeness of discussion when b =0, it gives$$x^2 = (-c/a)$$ - So 2 roots or no roots. (As discussed in my earlier post above).

St 2: |b|+|c|=0

This gives b = 0 and c = 0

So $$a(x+0)^2+0=0$$ -----> x =0

Only one possible values for x.

Sufficient.

Ans : B

Hope this helps!!
_________________

Everything will fall into place…

There is perfect timing for
everything and everyone.
Never doubt, But Work on
improving yourself,
Keep the faith and
It will all make sense.

Re: M60-12   [#permalink] 03 Oct 2018, 08:51
Display posts from previous: Sort by

# M60-12

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.