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Re: What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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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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Re: What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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09 Jul 2013, 15:25
What is the product of all the solutions of x^2  4x + 6 = 3  x  1 ? x^2  4x + 6 = 3  x  1 x^2  4x + 3 =  x  1
x>1 x^2  4x + 6 = 3  x  1 x^2  4x + 3 =  (x1) x^2  4x + 3 = x+1 x^2  3x +2 = 0 (x1)(x2)=0 x=1, x=2 2 falls within the range of x>1 x=2
x<1 x^2  4x + 6 = 3  x  1 x^2  4x + 6 = 3  (x1) x^2  4x + 6 = 3  (x+1) x^2  4x + 3 = + x  1 x^2  5x + 4 = 0 (x1)(x4) = 0 x=1, x=4 Neither 1 or 4 fall within the range of x<1 INVALID
The product is (2)
(C)
(Bunuel, could you explain to me how we know where the greater than or equals sign goes in these problems? I see that in your solution you had x>=1, but I don't know why that is as opposed to say, x<=1)
Thanks!



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Re: What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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19 Jun 2014, 02:57
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. Answer: C. Dear Bunuel, We do not need to consider two situation of x  1. As, x^2  4x + 6 = 3  x  1 <=> x^2  4x + 3 =  x  1 <0 => 1<x (<3) => x^2  4x + 3 = x1 => x^2  5x + 4 = 0 => 2 solutions



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What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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11 Aug 2015, 18:03
paul16 wrote: Hello everyone,
I am so close to understanding this question, but the one thing I do not understand is why the positive of x+2 is >= and the negative of x+2 is just <?
Sorry if its a dumb question
Paul I have this question too. Typically it doesnt matter if we use the +ve or ve absolute value function since the value is zero anyways. But for this case, we need to use the value to include/exclude values, so is >= and < the standard?



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Re: What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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11 Aug 2015, 18:15
ggurface wrote: paul16 wrote: Hello everyone,
I am so close to understanding this question, but the one thing I do not understand is why the positive of x+2 is >= and the negative of x+2 is just <?
Sorry if its a dumb question
Paul I have this question too. Typically it doesnt matter if we use the +ve or ve absolute value function since the value is zero anyways. But for this case, we need to use the value to include/exclude values, so is >= and < the standard? 1 thing, the text in red is not always true. x can be any value other than 0 as well. Coming back to your question, the reason we include "=" with ">" sign because x = x for x\(\geq\)0 while x =x for x<0 The convention is to always include "+" with ">" and not with "<" as the NATURE of x remains the same for x\(\geq\) 0, while the nature changes for x (we need to put a negative infront of 'x') when x<0. For example, 4 = 4 or 0 = 0 but 5 = (5) = 5 Hope this helps.



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Re: What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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10 Jun 2016, 17:54
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. Hi Bunuel, When I tried to follow the same method I get different answer for the below question. x^2 + 4x + 7 = x + 2 + 3. When x <= 0 then x + 2 =  (x+2 ) = x  2. Then x^2 + 4x + 7 =  x  2 + 3. => x^2 + 4x + 7 =  x + 1. => x^2 + 5x + 6 = 0 => ( x +2 ) ( x + 3) = 0 => x = 2 and  3 . ( Since x < = 0 both the numbers are possible ). When x > 0 then x + 2 = x + 2. Then x^2 + 4x + 7 = x + 2 + 3. => x^2 + 4x + 7 = x + 5. => x^2 + 3x + 2 = 0 . =>( x + 2 ) ( x + 1 ) = 0. x =  2 and  1 ( Since x > 0 both numbers doesn't fit in the range ). Then suitable numbers would be 2 and 3 and the product should be 6. Please clarify me if I am missing anything...



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Re: What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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11 Jun 2016, 00:43
msk0657 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. Hi Bunuel, When I tried to follow the same method I get different answer for the below question. x^2 + 4x + 7 = x + 2 + 3. When x <= 0 then x + 2 =  (x+2 ) = x  2. Then x^2 + 4x + 7 =  x  2 + 3. => x^2 + 4x + 7 =  x + 1. => x^2 + 5x + 6 = 0 => ( x +2 ) ( x + 3) = 0 => x = 2 and  3 . ( Since x < = 0 both the numbers are possible ). When x > 0 then x + 2 = x + 2. Then x^2 + 4x + 7 = x + 2 + 3. => x^2 + 4x + 7 = x + 5. => x^2 + 3x + 2 = 0 . =>( x + 2 ) ( x + 1 ) = 0. x =  2 and  1 ( Since x > 0 both numbers doesn't fit in the range ). Then suitable numbers would be 2 and 3 and the product should be 6. Please clarify me if I am missing anything... The transition point (value) must be the point for which the expression in modulus changes its sign. For x + 2 it's 2 not 0. So, you should consider the ranges when x<=2 and when x>2. Check highlighted text in my solution.
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Re: What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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25 Jul 2016, 20:52
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. Answer: C. Hi Bunuel, I don't know whether my alternative method is right to solve this question: 1) I tried to create a square: X^24X+6=3X1 X^24X+3=X1 X^22X+12X+2=X1 (X1)^22(X1)+X1=0 2) replace X1 with Y Y^22Y+Y=0 3) if Y>0, the equation will be: Y^2Y=0, Y(Y1)=0, then X1=0 or X1=1 So we have X=1 or X=2 4) if Y=0, X1=0, X=1 4) if Y<0, all in the left side are positive, thus cannot resulting in 0 Consequently, X=1 or X=2, the product is 2. Thanks,
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Re: What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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25 Jul 2016, 21:21
mod can be taken + and ve when taken +ve , solving the equation will give roots 1 and 2 , product will be 2 when taken ve , solving equation will give 1 and 4 , which can be discarded as we have considered x and ve and we are getting roots positive. answer will be 2



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Re: What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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11 Jan 2018, 10:34
I assume this is added by mistake into hard question. This is the easiest one.



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Re: What is the product of all the solutions of x^2  4x + 6=3 [#permalink]
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19 Jan 2018, 11:19
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 Edited the answer choice C from 2 to 2 and the OA. It's C not E. I just want to share my answer which could shorten calculation time a lot. x^2  4x + 6 = 3  x  1 X^2  4x + 3 =  x1 (X  3) (x  1) =  x  1 If x = 1, both side = 0 => 1 is a root If x not equal q, then: (X3) =   x  1 / (x1) If (x  1) > 0, (x 3 ) = 1 => x = 2. [2  1]>0 => 2 is a root If (x  1) < 0, (x 3) = 1 => x = 4. [41] >0. => 4 is not a root Conclusion: there are two root 1 and 2




Re: What is the product of all the solutions of x^2  4x + 6=3
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