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

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If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink] New post 07 Dec 2012, 05:30
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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
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Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink] New post 07 Dec 2012, 05:33
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Walkabout 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.

Answer: B.
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Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink] New post 10 Dec 2012, 06:53
which is the level of the question? 500?600?tks
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Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink] New post 13 Nov 2013, 11:31
Bunuel wrote:
Walkabout 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.

Answer: B.


Can you show how you manipulated both of the equations to get to zero, and to -1/2, and 3/2, thanks.
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Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink] New post 15 Nov 2013, 01:14
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selfishmofo wrote:
Bunuel wrote:
Walkabout 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.

Answer: B.


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.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. Tricky questions from previous years.

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink] New post 13 Apr 2014, 08:03
Bunuel wrote:
Walkabout 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.

Answer: B.


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?
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Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink] New post 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 \(= -\frac{1}{2}\)

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

Answer: B.


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!
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Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink] New post 19 May 2014, 23:41
1
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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.

Answer: B.


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.

Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm

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?
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

Hence answer is B.
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Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink] New post 06 Aug 2015, 18:51
Bunuel wrote:
Walkabout 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.

Answer: B.




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?
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Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x = [#permalink] New post 07 Aug 2015, 03:42
tigrr49 wrote:
Bunuel wrote:
Walkabout 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.

Answer: B.




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.
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Re: If x(2x + 1) = 0 and (x + 1/2)(2x - 3) = 0, then x =   [#permalink] 07 Aug 2015, 03:42
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