If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 23 Jan 2017, 13:30

### 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 June
Open Detailed Calendar

# If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =

Author Message
TAGS:

### Hide Tags

Manager
Joined: 02 Dec 2012
Posts: 178
Followers: 5

Kudos [?]: 2338 [0], given: 0

If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

07 Dec 2012, 05:30
6
This post was
BOOKMARKED
00:00

Difficulty:

5% (low)

Question Stats:

85% (01:56) correct 15% (01:07) wrong based on 953 sessions

### HideShow timer Statistics

If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =

(A) -3
(B) -1/2
(C) 0
(D) 1/2
(E) 3/2
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099

Kudos [?]: 93560 [2] , given: 10578

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

15 Nov 2013, 01:14
2
KUDOS
Expert's post
selfishmofo wrote:
Bunuel wrote:
If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =

(A) -3
(B) -1/2
(C) 0
(D) 1/2
(E) 3/2

x(2x + 1) = 0 --> x=0 OR x=-1/2;
(x + 1/2)(2x - 3) = 0 --> x=-1/2 OR x=3/2.

x=-1/2 satisfies both equations.

Can you show how you manipulated both of the equations to get to zero, and to -1/2, and 3/2, thanks.

When a product of two multiples is 0, it means that either of the multiples (or both) is 0.

$$x(2x + 1) = 0$$ --> $$x=0$$ or $$2x+1=0$$ ($$x=-\frac{1}{2}$$).

$$(x + \frac{1}{2})(2x - 3) = 0$$ --> $$x+\frac{1}{2}=0$$ ($$x=-\frac{1}{2}$$) or $$2x - 3=0$$ ($$x=\frac{3}{2}$$).

Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099

Kudos [?]: 93560 [1] , given: 10578

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

07 Dec 2012, 05:33
1
KUDOS
Expert's post
6
This post was
BOOKMARKED
If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =

(A) -3
(B) -1/2
(C) 0
(D) 1/2
(E) 3/2

$$x(2x + 1) = 0$$ --> $$x=0$$ OR $$x=-\frac{1}{2}$$;

$$(x + \frac{1}{2})(2x - 3) = 0$$ --> $$x=-\frac{1}{2}$$ OR $$x=\frac{3}{2}$$.

$$x=-\frac{1}{2}$$ satisfies both equations.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099

Kudos [?]: 93560 [1] , given: 10578

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

19 May 2014, 23:41
1
KUDOS
Expert's post
russ9 wrote:
russ9 wrote:
Bunuel wrote:

x(2x + 1) = 0 --> x=0 OR x=-1/2;
(x + 1/2)(2x - 3) = 0 --> x=-1/2 OR x=3/2.

x=-1/2 satisfies both equations.

This makes complete sense, although, I ran into trouble when I tried to FOIL the second equation and ended up with x^2-x-3/4=0 and from that point forward, I was completely stumped. Why is that method wrong?

I notice that I get confused on that front quite a bit - FOIL'ing vs. just setting both parenthesis to 0?

EDIT: As I was doing other problems, I ran into DS 67, Pg 180 of OG 13. The equation there is n(n+1) = 6, if I use the same methodology outlined above, the two solutions I get are n=6 and n=5. That is obviously wrong and I should've opted to FOIL in the above case. Hence my confusion -- why is it that in some situations I need to FOIL and in some other situations, I need to just equate the left to the right side WITHOUT foiling?

Hi Bunuel,

Still a little confused about the above question. I ran into countless more errors over the past few days -- cause was the same reason mentioned above. Would greatly appreciate some clarification.

Thanks!

When you have that the product of several multiples is equal to zero, then you don't need to expand the product. You can directly get the answer by equating these multiples to 0.

For example: $$(x - 3)(x + 2) = 0$$ --> $$x - 3 = 0$$ or $$x + 2 =0$$ --> $$x = 3$$ or $$x = -2$$. Now, you could expand and get $$x^2-x-6 = 0$$ and then solve with conventional method (check the links below) but the first approach is faster.

As for $$n(n+1) = 6$$. Don't know how you are getting the roots for this as 5 and 6, but for this question you cannot equate the multiples (n and n+1) to zero to get the roots because the product is not zero, it's 6. To solve it, you should expand to get: $$n^2+n-6=0$$ and then either solve by formula (check the links above) or by factoring to $$(n+3)(n-2)=0$$ and only then using the first approach: $$n + 3 = 0$$ or $$n - 2 =0$$ --> $$n = -3$$ or $$n = 2$$.

Does this make sense?
_________________
Intern
Joined: 12 Nov 2012
Posts: 7
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

10 Dec 2012, 06:53
which is the level of the question? 500?600?tks
Intern
Joined: 11 Aug 2013
Posts: 34
Followers: 0

Kudos [?]: 9 [0], given: 9

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

13 Nov 2013, 11:31
Bunuel wrote:
If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =

(A) -3
(B) -1/2
(C) 0
(D) 1/2
(E) 3/2

x(2x + 1) = 0 --> x=0 OR x=-1/2;
(x + 1/2)(2x - 3) = 0 --> x=-1/2 OR x=3/2.

x=-1/2 satisfies both equations.

Can you show how you manipulated both of the equations to get to zero, and to -1/2, and 3/2, thanks.
Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 53 [0], given: 23

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

13 Apr 2014, 08:03
Bunuel wrote:
If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =

(A) -3
(B) -1/2
(C) 0
(D) 1/2
(E) 3/2

x(2x + 1) = 0 --> x=0 OR x=-1/2;
(x + 1/2)(2x - 3) = 0 --> x=-1/2 OR x=3/2.

x=-1/2 satisfies both equations.

This makes complete sense, although, I ran into trouble when I tried to FOIL the second equation and ended up with x^2-x-3/4=0 and from that point forward, I was completely stumped. Why is that method wrong?

I notice that I get confused on that front quite a bit - FOIL'ing vs. just setting both parenthesis to 0?

EDIT: As I was doing other problems, I ran into DS 67, Pg 180 of OG 13. The equation there is n(n+1) = 6, if I use the same methodology outlined above, the two solutions I get are n=6 and n=5. That is obviously wrong and I should've opted to FOIL in the above case. Hence my confusion -- why is it that in some situations I need to FOIL and in some other situations, I need to just equate the left to the right side WITHOUT foiling?
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1858
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 47

Kudos [?]: 1937 [0], given: 193

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

22 Apr 2014, 01:55
$$(x + \frac{1}{2})(2x - 3) = 0$$

Multiply both sides by 2

(2x+1)(2x-3) = 0............ (1)

The other equation

x (2x+1) = 0 ............. (2)

With RHS 0 of both the equations, LHS has 2x+1 in common

Equating to 0

x $$= -\frac{1}{2}$$

_________________

Kindly press "+1 Kudos" to appreciate

Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 53 [0], given: 23

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

19 May 2014, 19:43
russ9 wrote:
Bunuel wrote:

x(2x + 1) = 0 --> x=0 OR x=-1/2;
(x + 1/2)(2x - 3) = 0 --> x=-1/2 OR x=3/2.

x=-1/2 satisfies both equations.

This makes complete sense, although, I ran into trouble when I tried to FOIL the second equation and ended up with x^2-x-3/4=0 and from that point forward, I was completely stumped. Why is that method wrong?

I notice that I get confused on that front quite a bit - FOIL'ing vs. just setting both parenthesis to 0?

EDIT: As I was doing other problems, I ran into DS 67, Pg 180 of OG 13. The equation there is n(n+1) = 6, if I use the same methodology outlined above, the two solutions I get are n=6 and n=5. That is obviously wrong and I should've opted to FOIL in the above case. Hence my confusion -- why is it that in some situations I need to FOIL and in some other situations, I need to just equate the left to the right side WITHOUT foiling?

Hi Bunuel,

Still a little confused about the above question. I ran into countless more errors over the past few days -- cause was the same reason mentioned above. Would greatly appreciate some clarification.

Thanks!
Intern
Joined: 14 May 2014
Posts: 45
Followers: 0

Kudos [?]: 40 [0], given: 1

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

20 May 2014, 08:20
Multiplication of any number with 0 gives 0 ..

When product of two or more expressions is zero, then any expression or product of combination of expressions can be 0.

Here,
x(2x+1) = 0
here product of x and 2x+1 equals 0 => either x = 0 or 2x+1 = 0 or both

hence x=0 or -1/2 will satisfy the first equation.

(x+1/2)(2x-3) = 0
here product of x+1/2 and 2x-3 equals 0=> either x+1/2 = 0 or 2x-3 = 0 or both

hence x = -1/2 or 3/2 will satisfy the second equation.
-1/2 is common in both sets hence x= -1/2 will satisfy both equations.

_________________

Help me with Kudos if it helped you "

Mathematics is a thought process.

Intern
Joined: 18 May 2014
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 2

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

06 Aug 2015, 18:51
Bunuel wrote:
If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =

(A) -3
(B) -1/2
(C) 0
(D) 1/2
(E) 3/2

$$x(2x + 1) = 0$$ --> $$x=0$$ OR $$x=-\frac{1}{2}$$;

$$(x + \frac{1}{2})(2x - 3) = 0$$ --> $$x=-\frac{1}{2}$$ OR $$x=\frac{3}{2}$$.

$$x=-\frac{1}{2}$$ satisfies both equations.

Should we apply the same logic to data sufficiency questions if we are asked to find the value of x and given each quadratic as one of the statements? Thus, statement 1 and statement 2 would each be insufficient but combining together they are sufficient (i.e. the answer is C) due to one shared solution for both quadratics/statements?
Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2654
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Followers: 117

Kudos [?]: 1342 [0], given: 789

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

07 Aug 2015, 03:42
tigrr49 wrote:
Bunuel wrote:
If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =

(A) -3
(B) -1/2
(C) 0
(D) 1/2
(E) 3/2

$$x(2x + 1) = 0$$ --> $$x=0$$ OR $$x=-\frac{1}{2}$$;

$$(x + \frac{1}{2})(2x - 3) = 0$$ --> $$x=-\frac{1}{2}$$ OR $$x=\frac{3}{2}$$.

$$x=-\frac{1}{2}$$ satisfies both equations.

Should we apply the same logic to data sufficiency questions if we are asked to find the value of x and given each quadratic as one of the statements? Thus, statement 1 and statement 2 would each be insufficient but combining together they are sufficient (i.e. the answer is C) due to one shared solution for both quadratics/statements?

Technically, yes, C would be your answer in a similarly worded DS problem. But beware that not all quadratic equations give you 2 distinct values.

Lets say, statement 1 gave you:

$$x^2+2x+1$$ = 0 , this is in fact $$(x+1)^2$$ ---> $$(x+1)^2 = 0$$---> $$x = -1$$ .

So you get ONLY 1 value and thus this statement will be sufficient on its own. Thus, in DS questions, you should be making sure that you actually are getting 2 distinct values.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Manager
Joined: 06 Jun 2014
Posts: 58
Followers: 0

Kudos [?]: 8 [0], given: 105

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

26 Jan 2016, 16:35
I do understant the Method given by Bunuel, but I tried to look for different solution and came to the following :

Equation 1 : x(2x+1)=0 => 2x^2 +x=0 => 2x^2=-x

Equation 2: ( X+1/2)(2x+3) after FOIL => 2x^2 -2x-3/2 , and here I replaced 2x^2 with -x from the first equation and got
-x-2x-3/2=o => -3x=3/2 => x=-1/2

I would like kindly ask Bunuel or other experts if this is also a correct solution?
Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2654
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Followers: 117

Kudos [?]: 1342 [0], given: 789

If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

26 Jan 2016, 16:42
kzivrev wrote:
I do understant the Method given by Bunuel, but I tried to look for different solution and came to the following :

Equation 1 : x(2x+1)=0 => 2x^2 +x=0 => 2x^2=-x

Equation 2: ( X+1/2)(2x+3) after FOIL => 2x^2 -2x-3/2 , and here I replaced 2x^2 with -x from the first equation and got
-x-2x-3/2=o => -3x=3/2 => x=-1/2

I would like kindly ask Bunuel or other experts if this is also a correct solution?

Yes, this is a correct method but a bit more time consuming method. In GMAT, you must do proper time management and not just solve the questions correctly.

When you are given a*b =0 ---> 3 cases possible

1. a=0 and b $$neq$$ 0

2. b=0 and a $$neq$$ 0

3. a=b=0

When you are directly given x(2x + 1) = 0 --> either x=0 or 2x+1 =0 --> x=-0.5.

Similarly from the second equation, (x + 1/2)(2x - 3) = 0 ---> either x+0.5 =0 --> x=0 or 2x-3 =0 -->x=1.5.

From these 2 sets of solutions, you see that x=-0.5 is the common solution and is hence the value of x asked in the question.

Hope this helps.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Director
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 516
Location: United States (CA)
Followers: 20

Kudos [?]: 195 [0], given: 2

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink]

### Show Tags

09 Jun 2016, 13:03
If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =

(A) -3
(B) -1/2
(C) 0
(D) 1/2
(E) 3/2

To solve we will use the zero product property. The zero product property states that if the product of two quantities is equal to 0, then at least one of the quantities has to be equal to 0. That is, if a  b = 0, then either a = 0 or b = 0. Of course, both a and b can be 0 at the same time. The point is that at least one of them has to be 0.

Let’s start determining the value(s) of x in the equation x(2x + 1) = 0

If x(2x + 1) = 0, we know:

x = 0

OR

2x + 1 = 0

2x = -1

x = -1/2

Thus, x = 0 or x = -1/2

Let’s now determine the value(s) of x in the second equation (x + 1/2)(2x - 3) = 0

(x + 1/2)(2x - 3) = 0, we know:

(x + 1/2) = 0

x = -1/2

OR

(2x - 3) = 0

2x = 3

x = 3/2

Thus, x = -1/2 or x = 3/2

Because we need to determine a value for x in both equations, the answer is x = -1/2.

_________________

Jeffrey Miller
Scott Woodbury-Stewart
Founder and CEO

Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =   [#permalink] 09 Jun 2016, 13:03
Similar topics Replies Last post
Similar
Topics:
2 How to solve |x|^2 - x >0 6 06 Jun 2016, 08:41
1 If x ≠ 0 and x - (2- x^2)/x = y/x, then y = 4 13 Dec 2015, 05:12
5 If 'a' and 'b' are the roots of the quadratic equation x^2 - x + 1 = 0 3 12 Sep 2015, 10:14
82 If x#0, then root(x^2)/x= 25 14 Dec 2009, 16:58
12 If x#0, then root(x^2)/x= 3 18 May 2008, 12:29
Display posts from previous: Sort by