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What is the product of all the solutions of x^2  4x + 6=3
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What is the product of all the solutions of \(x^2  4x + 6 = 3  x  1\) ? (A) 8 (B) 4 (C) 2 (D) 4 (E) 8 Edited the answer choice C from 2 to 2 and the OA. It's C not E.
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Originally posted by carcass on 13 Feb 2013, 05:22.
Last edited by carcass on 04 Jul 2018, 09:59, edited 3 times in total.
Edited the answer choices and the OA.




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Re: What is the product of all the solutions of x^2  4x + 6=3
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13 Feb 2013, 10:13
carcass wrote: What is the product of all the solutions of x^2  4x + 6 = 3  x  1 ?
(A) 8 (B) 4 (C) 2 (D) 4 (E) 8 If \(x<1\), then \(x  1 = (x1)=1x\), so in this case we'll have \(x^2  4x + 6 = 3(1x)\) > \(x^25x+4=0\) > \(x=1\) or \(x=4\) > discard both solutions since neither is in the range \(x<1\). If \(x\geq{1}\), then \(x  1 = x1\), so in this case we'll have \(x^2  4x + 6 = 3(x1)\) > \(x^23x+2=0\) > \(x=1\) or \(x=2\). Therefore, the product of the roots is 1*2=2. Answer: C.
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Re: What is the product of all the solutions of x^2  4x + 6=3
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13 Feb 2013, 05:28
carcass wrote: What is the product of all the solutions of \(x^2  4x + 6 = 3  x  1\)? (A) 8 (B) 4 (C) 2 (D) 4 (E) 8 If x  1>=0 >then the modulus will be equal to (x1) & roots of the resulting equation will be 2,1 If x  1<0 >then the modulus will be equal to (x+1) & roots of the resulting equation will be 4,1 So the product of all the roots (2,1,4,1) is 8.
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Re: What is the product of all the solutions of x^2  4x + 6=3
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13 Feb 2013, 09:14
In order for x  1 to be equal to 1  x, we would have to have x < 1 . Therefore eliminating your second pair of solutions
You could also verify this by substituting x = 4 inthe original equation, and seeing that this solution DOES NOT fit. The only two solutions are 1 and 2.
ANS: no correct option available Posted from my mobile device



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Re: What is the product of all the solutions of x^2  4x + 6=3
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14 Feb 2013, 11:44
Bunuel wrote: carcass wrote: What is the product of all the solutions of x^2  4x + 6 = 3  x  1 ?
(A) 8 (B) 4 (C) 2 (D) 4 (E) 8 If \(x<1\), then \(x  1 = (x1)=1x\), so in this case we'll have \(x^2  4x + 6 = 3(1x)\) > \(x^25x+4=0\) > \(x=1\) or \(x=4\) > discard both solutions since neither is in the range \(x<1\). If \(x\geq{1}\), then \(x  1 = x1\), so in this case we'll have \(x^2  4x + 6 = 3(x1)\) > \(x^23x+2=0\) > \(x=1\) or \(x=2\). Therefore, the product of the roots is 1*2=2. No correct answer among the choices. Wow! Really? How often do we see this? Who the heck wrote this question?



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Re: What is the product of all the solutions of x^2  4x + 6=3
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15 Feb 2013, 03:43
vandygrad11 wrote: Bunuel wrote: carcass wrote: What is the product of all the solutions of x^2  4x + 6 = 3  x  1 ?
(A) 8 (B) 4 (C) 2 (D) 4 (E) 8 If \(x<1\), then \(x  1 = (x1)=1x\), so in this case we'll have \(x^2  4x + 6 = 3(1x)\) > \(x^25x+4=0\) > \(x=1\) or \(x=4\) > discard both solutions since neither is in the range \(x<1\). If \(x\geq{1}\), then \(x  1 = x1\), so in this case we'll have \(x^2  4x + 6 = 3(x1)\) > \(x^23x+2=0\) > \(x=1\) or \(x=2\). Therefore, the product of the roots is 1*2=2. No correct answer among the choices. Wow! Really? How often do we see this? Who the heck wrote this question? How often do we see what? I've seen similar question which reads: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 ?A. 6 B. 2 C. 2 D. 6 E. 12 OA: Try it. I'll provide solution for this question later, if necessary.
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Re: What is the product of all the solutions of x^2  4x + 6=3
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16 Feb 2013, 12:37
Bunuel wrote: How often do we see what? I've seen similar question which reads: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 ?A. 6 B. 2 C. 2 D. 6 E. 12 OA: Try it. I'll provide solution for this question later, if necessary. How often do we see no correct answer among the answer choices?



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Re: What is the product of all the solutions of x^2  4x + 6=3
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18 Feb 2013, 04:27
Bunuel wrote: carcass wrote: What is the product of all the solutions of x^2  4x + 6 = 3  x  1 ?
(A) 8 (B) 4 (C) 2 (D) 4 (E) 8 If \(x<1\), then \(x  1 = (x1)=1x\), so in this case we'll have \(x^2  4x + 6 = 3(1x)\) > \(x^25x+4=0\) > \(x=1\) or \(x=4\) > discard both solutions since neither is in the range \(x<1\). If \(x\geq{1}\), then \(x  1 = x1\), so in this case we'll have \(x^2  4x + 6 = 3(x1)\) > \(x^23x+2=0\) > \(x=1\) or \(x=2\). Therefore, the product of the roots is 1*2=2. No correct answer among the choices. I have a question, I do understand that why have you taken the value 1 but I don't understand why have you taken x>=1. Why not simply x>1



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Re: What is the product of all the solutions of x^2  4x + 6=3
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18 Feb 2013, 04:35
davidfrank wrote: Bunuel wrote: carcass wrote: What is the product of all the solutions of x^2  4x + 6 = 3  x  1 ?
(A) 8 (B) 4 (C) 2 (D) 4 (E) 8 If \(x<1\), then \(x  1 = (x1)=1x\), so in this case we'll have \(x^2  4x + 6 = 3(1x)\) > \(x^25x+4=0\) > \(x=1\) or \(x=4\) > discard both solutions since neither is in the range \(x<1\). If \(x\geq{1}\), then \(x  1 = x1\), so in this case we'll have \(x^2  4x + 6 = 3(x1)\) > \(x^23x+2=0\) > \(x=1\) or \(x=2\). Therefore, the product of the roots is 1*2=2. No correct answer among the choices. I have a question, I do understand that why have you taken the value 1 but I don't understand why have you taken x>=1. Why not simply x>1 x could be 1, thus when you consider the ranges you should include this value in either of the range, so we could consider x<1 and x>=1 OR x<=1 and x>1 (you cam include = sign in either of the ranges). Hope it's clear.
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Re: What is the product of all the solutions of x^2  4x + 6=3
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18 Feb 2013, 04:48
vandygrad11 wrote: Bunuel wrote: How often do we see what? I've seen similar question which reads: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 ?A. 6 B. 2 C. 2 D. 6 E. 12 OA: Try it. I'll provide solution for this question later, if necessary. How often do we see no correct answer among the answer choices? Never, if it's a proper GMAT question.
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Re: What is the product of all the solutions of x^2  4x + 6=3
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Updated on: 18 Feb 2013, 06:47
I've seen similar question which reads: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 ?A. 6 B. 2 C. 2 D. 6 E. 12 OA: Try it. I'll provide solution for this question later, if necessary.[/quote] if x+2>=0 then x+2= (x+2) eqation becomes (x+2)(x+1)=0 x=2,1 if x+2<0 then x+2=(x+2) equation becomes (x+2) (x+3) =0 x=2,3 ( can't be 2 since x<2) product of the solution  2*1*3= 6 Ans
Originally posted by guerrero25 on 18 Feb 2013, 06:16.
Last edited by guerrero25 on 18 Feb 2013, 06:47, edited 1 time in total.



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Re: What is the product of all the solutions of x^2  4x + 6=3
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18 Feb 2013, 06:32
guerrero25 wrote: I've seen similar question which reads: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 ?A. 6 B. 2 C. 2 D. 6 E. 12 OA: Try it. I'll provide solution for this question later, if necessary. if x+2>=2 then x+2= (x+2) eqation becomes (x+2)(x+1)=0 x=2,1 if x+2<2 then x+2=(x+2) equation becomes (x+2) (x+3) =0 x=2,3 ( can't be 2 since x<2) product of the solution  2*1*3= 6 Ans I guess you meant the following: When \(x\leq{2}\), then \(x+2=(x2)\). When \(x>{2}\), then \(x+2=(x2)\). Complete solution: \(x^2 + 4x + 7 = x + 2 + 3\) > \(x^2 + 4x + 4 = x + 2\) > \((x+2)^2=x+2\) > \((x+2)^4=(x+2)^2\) > \((x+2)^2((x+2)^21)=0\): \(x+2=0\) > \(x=2\); OR \((x+2)^21=0\) > \((x+2)^2=1\) > \(x=1\) or \(x=3\). The product of the roots: \((2)*(1)*(3)=6\). Answer: A. Hope it's clear.
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Re: What is the product of all the solutions of x^2  4x + 6=3
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18 Feb 2013, 06:48
Bunuel wrote: guerrero25 wrote: I've seen similar question which reads: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 ?A. 6 B. 2 C. 2 D. 6 E. 12 OA: Try it. I'll provide solution for this question later, if necessary. if x+2>=2 then x+2= (x+2) eqation becomes (x+2)(x+1)=0 x=2,1 if x+2<2 then x+2=(x+2) equation becomes (x+2) (x+3) =0 x=2,3 ( can't be 2 since x<2) product of the solution  2*1*3= 6 Ans I guess you meant the following: When \(x\leq{2}\), then \(x+2=(x2)\). When \(x>{2}\), then \(x+2=(x2)\). Complete solution: \(x^2 + 4x + 7 = x + 2 + 3\) > \(x^2 + 4x + 4 = x + 2\) > \((x+2)^2=x+2\) > \((x+2)^4=(x+2)^2\) > \((x+2)^2((x+2)^21)=0\): \(x+2=0\) > \(x=2\); OR \((x+2)^21=0\) > \((x+2)^2=1\) > \(x=1\) or \(x=3\). The product of the roots: \((2)*(1)*(3)=6\). Answer: A. Hope it's clear. thanks ! That was a Typo . I edited the post .



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Re: What is the product of all the solutions of x^2  4x + 6=3
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18 Feb 2013, 06:52
guerrero25 wrote: I've seen similar question which reads: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 ?A. 6 B. 2 C. 2 D. 6 E. 12 OA: Try it. I'll provide solution for this question later, if necessary. if x+2>=0 then x+2= (x+2) eqation becomes (x+2)(x+1)=0 x=2,1 if x+2<0 then x+2=(x+2) equation becomes (x+2) (x+3) =0 x=2,3 ( can't be 2 since x<2) product of the solution  2*1*3= 6 Ans[/quote] Still not correct. Absolute value cannot be negative, so x+2 is always more than or equal to zero.
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Re: What is the product of all the solutions of x^2  4x + 6=3
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05 Jul 2013, 13:27
Bunuel wrote: guerrero25 wrote: I've seen similar question which reads: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 ?A. 6 B. 2 C. 2 D. 6 E. 12 OA: Try it. I'll provide solution for this question later, if necessary. if x+2>=2 then x+2= (x+2) eqation becomes (x+2)(x+1)=0 x=2,1 if x+2<2 then x+2=(x+2) equation becomes (x+2) (x+3) =0 x=2,3 ( can't be 2 since x<2) product of the solution  2*1*3= 6 Ans I guess you meant the following: When \(x\leq{2}\), then \(x+2=(x2)\). When \(x>{2}\), then \(x+2=(x2)\). Complete solution: \(x^2 + 4x + 7 = x + 2 + 3\) > \(x^2 + 4x + 4 = x + 2\) > \((x+2)^2=x+2\) > \((x+2)^4=(x+2)^2\) > \((x+2)^2((x+2)^21)=0\): \(x+2=0\) > \(x=2\); OR \((x+2)^21=0\) > \((x+2)^2=1\) > \(x=1\) or \(x=3\). The product of the roots: \((2)*(1)*(3)=6\). Answer: A. Hope it's clear. Bunuel, Can you solve this problem using the other method that you used in the previous problem? I mean: If x >= 0, x + 2 = x + 2. This would give the equation: x^2 + 4x + 7 = x + 5. Roots are 2, and 1 what is the other scenario? What happens if x < 0? How do we end up with the roots 3, and 1?? Thaanks



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Re: What is the product of all the solutions of x^2  4x + 6=3
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05 Jul 2013, 13:40
jjack0310 wrote: Bunuel wrote: guerrero25 wrote: I've seen similar question which reads: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 ?A. 6 B. 2 C. 2 D. 6 E. 12 OA: Try it. I'll provide solution for this question later, if necessary. if x+2>=2 then x+2= (x+2) eqation becomes (x+2)(x+1)=0 x=2,1 if x+2<2 then x+2=(x+2) equation becomes (x+2) (x+3) =0 x=2,3 ( can't be 2 since x<2) product of the solution  2*1*3= 6 Ans I guess you meant the following: When \(x\leq{2}\), then \(x+2=(x2)\). When \(x>{2}\), then \(x+2=(x2)\). Complete solution: \(x^2 + 4x + 7 = x + 2 + 3\) > \(x^2 + 4x + 4 = x + 2\) > \((x+2)^2=x+2\) > \((x+2)^4=(x+2)^2\) > \((x+2)^2((x+2)^21)=0\): \(x+2=0\) > \(x=2\); OR \((x+2)^21=0\) > \((x+2)^2=1\) > \(x=1\) or \(x=3\). The product of the roots: \((2)*(1)*(3)=6\). Answer: A. Hope it's clear. Bunuel, Can you solve this problem using the other method that you used in the previous problem? I mean: If x >= 0, x + 2 = x + 2. This would give the equation: x^2 + 4x + 7 = x + 5. Roots are 2, and 1 what is the other scenario? What happens if x < 0? How do we end up with the roots 3, and 1?? Thaanks When \(x\leq{2}\), then \(x+2=(x2)\). So, in this case we'll have \(x^2 + 4x + 7 =(x + 2) + 3\) > \(x=3\) or \(x=2\). Both solutions are valid. When \(x>{2}\), then \(x+2=(x2)\). So, in this case we'll have \(x^2 + 4x + 7 =(x + 2) + 3\) > \(x=2\) or \(x=1\). The first solution is not valid since it's out of the range we consider. The second one is OK. So, there are 3 valid solutions: \(x=3\), \(x=2\) and \(x=1\). Hope it's clear.
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Re: What is the product of all the solutions of x^2  4x + 6=3
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06 Jul 2013, 09:09
Sorry. It is not clear.
Can you explain what was wrong with the way I was approaching the problem?
I mean other than the part that you marked red, what was I doing wrong? Do I have to solve theproblem using the solution that you mentioned?
If x > 2, how is x + 2 = (x  2)? Is there an identity that I am missing? If I plug in, X = 1, x + 2 = 1, but (x  2) = 3
Why the discrepancy? What identity am I missing?



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Re: What is the product of all the solutions of x^2  4x + 6=3
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06 Jul 2013, 09:16
jjack0310 wrote: Sorry. It is not clear.
Can you explain what was wrong with the way I was approaching the problem?
I mean other than the part that you marked red, what was I doing wrong? Do I have to solve theproblem using the solution that you mentioned?
If x > 2, how is x + 2 = (x  2)? Is there an identity that I am missing? If I plug in, X = 1, x + 2 = 1, but (x  2) = 3
Why the discrepancy? What identity am I missing? There was a typo: When \(x\leq{2}\), then \(x+2=(x+2)\) When \(x>{2}\), then \(x+2=(x+2)\). Absolute value properties:When \(x\leq {0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x \geq {0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). For our question, when x>2 (when x+2>0), x+2=x+2. Hope it's clear.
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Re: What is the product of all the solutions of x^2  4x + 6=3
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06 Jul 2013, 09:52
Bunuel wrote: jjack0310 wrote: Sorry. It is not clear.
Can you explain what was wrong with the way I was approaching the problem?
I mean other than the part that you marked red, what was I doing wrong? Do I have to solve theproblem using the solution that you mentioned?
If x > 2, how is x + 2 = (x  2)? Is there an identity that I am missing? If I plug in, X = 1, x + 2 = 1, but (x  2) = 3
Why the discrepancy? What identity am I missing? There was a typo: When \(x\leq{2}\), then \(x+2=(x+2)\) When \(x>{2}\), then \(x+2=(x+2)\). Absolute value properties:When \(x \leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x \geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). For our question, when x>2 (when x+2>0), x+2=x+2. Hope it's clear. Got it. Thanks, Final question, why are there two possibilities for when x = 0? Is that correct? or a typo?



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Re: What is the product of all the solutions of x^2  4x + 6=3
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06 Jul 2013, 09:53
jjack0310 wrote: Bunuel wrote: jjack0310 wrote: Sorry. It is not clear.
Can you explain what was wrong with the way I was approaching the problem?
I mean other than the part that you marked red, what was I doing wrong? Do I have to solve theproblem using the solution that you mentioned?
If x > 2, how is x + 2 = (x  2)? Is there an identity that I am missing? If I plug in, X = 1, x + 2 = 1, but (x  2) = 3
Why the discrepancy? What identity am I missing? There was a typo: When \(x\leq{2}\), then \(x+2=(x+2)\) When \(x>{2}\), then \(x+2=(x+2)\). Absolute value properties:When \(x \leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x \geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). For our question, when x>2 (when x+2>0), x+2=x+2. Hope it's clear. Got it. Thanks, Final question, why are there two possibilities for when x = 0? Is that correct? or a typo? No, that's not a typo: 0=0=0.
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