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Sub 505 (Easy)|   Algebra|                                          
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Walkabout
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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Bunuel
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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

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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Bunuel
Walkabout
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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Bunuel


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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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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Walkabout
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.

The answer is B.
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Hi all ,

I ended up by solving the two quadratic equations simultaneously to find the value of x instead of solving one by one for the value of x .

Posted from my mobile device
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LeenaSai
Hi all ,

I ended up by solving the two quadratic equations simultaneously to find the value of x instead of solving one by one for the value of x .

Posted from my mobile device

LeenaSai

You don’t need to solve for quadratic here. It’s already given in solved form.

If a quadratic equation after factorisation would look like (x - a)(x - b) = 0
—> x - a = 0 or x - b = 0
—> x = a or b

Given 1st quadratic equation,
x(2x + 1) = 0
—> x = 0 or 2x + 1 = 0
—> x = 0 or -1/2

and 2nd quadratic equation,
(x + 1/2)(2x - 3) = 0
—> x + 1/2 = 0 or 2x - 3 = 0
—> x = -1/2 or x = 3/2

Combining 1st and 2nd equation, we get common value of x as -1/2

So, Option B
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Hi All,

This question gives us a couple of Quadratic equations that have already been factored:

(X)(2X + 1) = 0 and (X + ½)(2X – 3) = 0

We’re asked when X equals. This is a standard mid-level Algebra prompt that has already done half the work for us, so we just need to calculate the two values of X in each equation and find the one answer that shows up in BOTH equations.

With (X)(2X + 1) = 0

X = 0 and X = -1/2 are the two solutions

With (X + ½)(2X – 3) = 0

X = -1/2 and X = 3/2 are the two solutions

Based on these results, the answer that shows up in BOTH is…

Final Answer:
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Rich
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Walkabout
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

Answer: Option B

Step-by-Step Video solution by GMATinsight

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I just had this question on in the GMAT OG Practice questions. I dont know if anyone else has encountered this, but I want to make sure I am not crazy. See attached picture.

The prompt gives:

x(2x+1)=0 and (X+1/2)(2x-3)=0

In this problem there are two different x variables. Big X and little x. I took this to mean that they were completely separate variables. Then the question asks what little x can equal, -1/2 cannot be right because in the second equation, little x can never be -1/2. Is this a typo on GMAC's part? Or am I always supposed to assume that big X and little x are the same variable? I know 0 is not the right answer, but i just picked something because it wasnt making sense to me.
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Bunuel
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jafeezy
I just had this question on in the GMAT OG Practice questions. I dont know if anyone else has encountered this, but I want to make sure I am not crazy. See attached picture.

The prompt gives:

x(2x+1)=0 and (X+1/2)(2x-3)=0

In this problem there are two different x variables. Big X and little x. I took this to mean that they were completely separate variables. Then the question asks what little x can equal, -1/2 cannot be right because in the second equation, little x can never be -1/2. Is this a typo on GMAC's part? Or am I always supposed to assume that big X and little x are the same variable? I know 0 is not the right answer, but i just picked something because it wasnt making sense to me.

That's a typo. Here is a screenshot from the OG:


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