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

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

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

1

This post received 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.

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.

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 \(n = 2\).

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?

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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