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What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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21 Dec 2014, 21:17
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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 My solution looked like this:
If x<0, then x^2+4x+7=(x+2)+3 >x^2+4x+7=x2+3> (x+2)(x+3)=0 >x=2;3
If x>=0, then x^2+4x+7=(x+2)+3>x^2+4x+7=x+2+3>(x+2)(x+1)=0 >x=2,1, neither of the solutions are valid because for this case x>=0.
So, for my answer I got 6, which was incorrect. Obviously, my mistake was here: "x=2,1 >neither of the solutions are valid because for this case x>=0", could anybody explain to me why am I wrong?
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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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21 Dec 2014, 23:17
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viktorija wrote: 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
My solution looked like this:
If x<0, then x^2+4x+7=(x+2)+3 >x^2+4x+7=x2+3> (x+2)(x+3)=0 >x=2;3
If x>=0, then x^2+4x+7=(x+2)+3>x^2+4x+7=x+2+3>(x+2)(x+1)=0 >x=2,1, neither of the solutions are valid because for this case x>=0.
So, for my answer I got 6, which was incorrect. Obviously, my mistake was here: "x=2,1 >neither of the solutions are valid because for this case x>=0", could anybody explain to me why am I wrong? there will be two cases . 1. wen mod of (x+2)<0 x<2 So (x+2)(x+3)=0; x= 3 2. Wen mod of (x+2)>=0 x>=2 so (x+2)(x+1)=0; x can take both values 1 &2 combining both cases product will be 1*2*3=6



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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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22 Dec 2014, 00:02
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x^2+4x+7=x+2+3
x^2+4x+4=x+2
(X+2)^2 = x+2 Only 3 values possible, x = 3, x = 2, x =1 So multiplication = 6



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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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22 Dec 2014, 02:53
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viktorija wrote: My solution looked like this:
If x<0, then x^2+4x+7=(x+2)+3 >x^2+4x+7=x2+3> (x+2)(x+3)=0 >x=2;3
for x<0, possible values of x are 1,2,3....... now. for x=1, x+2 will be positive. hence the range you have selected for this equation is wrong. it should be x<2 Quote: If x>=0, then x^2+4x+7=(x+2)+3>x^2+4x+7=x+2+3>(x+2)(x+1)=0 >x=2,1, neither of the solutions are valid because for this case x>=0. by taking x>0. you have neglected the case of x=1 for which x+2 will be positive. thus your desired range should be x>2. you can read more about the modulus functions here. http://gmatclub.com/forum/mathabsolutevaluemodulus86462.html



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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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22 Dec 2014, 03:26
viktorija wrote: 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
My solution looked like this:
If x<0, then x^2+4x+7=(x+2)+3 >x^2+4x+7=x2+3> (x+2)(x+3)=0 >x=2;3
If x>=0, then x^2+4x+7=(x+2)+3>x^2+4x+7=x+2+3>(x+2)(x+1)=0 >x=2,1, neither of the solutions are valid because for this case x>=0.
So, for my answer I got 6, which was incorrect. Obviously, my mistake was here: "x=2,1 >neither of the solutions are valid because for this case x>=0", could anybody explain to me why am I wrong? Similar question to practice: whatistheproductofallthesolutionsofx24x147152.html
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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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22 Dec 2014, 05:12
Two cases for the RHS viz (x+5) and (x+1). Equating to LHS gives sols as 1, 2, 3 Product = 6



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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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22 Dec 2014, 12:49
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manpreetsingh86 wrote: viktorija wrote: My solution looked like this:
If x<0, then x^2+4x+7=(x+2)+3 >x^2+4x+7=x2+3> (x+2)(x+3)=0 >x=2;3
for x<0, possible values of x are 1,2,3....... now. for x=1, x+2 will be positive. hence the range you have selected for this equation is wrong. it should be x<2 Quote: If x>=0, then x^2+4x+7=(x+2)+3>x^2+4x+7=x+2+3>(x+2)(x+1)=0 >x=2,1, neither of the solutions are valid because for this case x>=0. by taking x>0. you have neglected the case of x=1 for which x+2 will be positive. thus your desired range should be x>2. you can read more about the modulus functions here. http://gmatclub.com/forum/mathabsolutevaluemodulus86462.htmlThank you, manpreetsingh86, for the explanation! Now I see where my mistake was.



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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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22 Dec 2014, 14:51
Hi viktorija, The "tricky" part of this question is the absolute value. When that type of math is involved, Test Takers often miss out on some of the possible answers, so it's important to do whatever you have to do to be sure that you've found all of the possibilities. To that end, I'm going to use ALL of the little clues that question gives me and avoid as much "crazy" math as possible. We're given X^2+4X+7=X+2+3 and we asked for the product of ALL of the solutions. Notice how the question asks us to find ALL of the solutions and NOT "both" solutions. This gets me thinking that there's probably more than 2 solutions. Also, quadratic equations usually have 1 or 2 solutions and absolute values usually have 1 or 2 solutions, so I'd be looking for up to 3 or 4 possibilities.... First, we simplify: X^2+4X+7 = X+2+3 X^2+4X+4 = X+2 (X+2)(X+2) = X+2 Now, looking at that last line, there's NO WAY that x can be positive. Try TESTing any positive number  the two sides will NEVER equal. So there are NO positive answers. Looking at the answer choices, they're all within a relatively small distance from 0, so the individual values for X are probably all relatively close to 0 as well. Let's find them with a bit of "brute force"... X = 0 (0+2)(0+2) = 0+2? (2)(2) = 2? NO, 0 is NOT a solution X = 1 (1+2)(1+2) = 1+2 (1)(1) = 1? YES, 1 IS a solution X = 2 (2+2)(2+2) = 2+2 (0)(0) = 0? YES, 2 IS a solution Remember that there were probably going to be MORE than 2 solutions, so we have to keep looking... X = 3 (3+2)(3+2) = 3+2 (1)(1) = 1? YES, 3 IS a solution At this point, if we take the product of these 3 answers we have (1)(2)(3) = 6. Looking at the 5 answer choices, there CAN'T be any other answers. To go from 6 to 12, (2) would have to be an answer (but we already know that positive numbers DON'T FIT). To go from 6 to +12, (2) would have to be used twice, which is not a mathematical option. So 6 MUST be the answer. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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14 Sep 2017, 05:46
I missed the (1) root  silly mistake.
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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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16 Sep 2017, 07:44
viktorija wrote: 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 Let’s first solve for when x + 2 is positive: 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 or x = 1 Next we can solve for when x + 2 is negative: x^2 + 4x + 7 = (x + 2) + 3 x^2 + 4x + 7 = x + 1 x^2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x = 3 or x = 2 We see that the product of all the solutions is (2)(1)(3) = 6. Answer: A
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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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13 Feb 2018, 16:05
If solving for x2 gives the solutions x=3 and x=2, and solving for x+2 gives the solutions x=2 and x=1 why does the answer take into account only 3*2*1 = 6 when the question asks for "all solutions"? If it is asking for all solutions why are we not taking 3*2*1*2 = 12 ?
From the discussions and solutions above, I tried to to rationalize this way:
1) x+2>0 so x>2
i.e x^2+4x+7=x+2+3 results: x=2 and 1
checking: 2+2>0 0>0 {null so doesn't apply}
1+2>0 1>0 {ok}
2) (x2)>0 so x<2 i.e x^2+4x+7=x2+3 results x = 2 and 3
checking: (2)+2>0 4>0 {ok}
(3)+2>0 5>0 {ok}
Am I on the right page? I am seeking validation for comprehending the solution of 6 versus the incorrect solution of 12. Do I treat the repeated value as one solution? I ask this so if in the future this type of question provides 4 distinct solutions (no repeated value), would we multiply all 4 solutions or do a further check to see the solution holds true against the inequalities?



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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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19 Feb 2018, 09:48
khan0210 wrote: If solving for x2 gives the solutions x=3 and x=2, and solving for x+2 gives the solutions x=2 and x=1 why does the answer take into account only 3*2*1 = 6 when the question asks for "all solutions"? If it is asking for all solutions why are we not taking 3*2*1*2 = 12 ? I have the exact same doubt. From what i get, people are doing multiplication of 3, 2, and 1. So, 2 is not being multiplied twice. Not sure why. Perhaps the question should have asked for product of "distinct" solutions of x. Can someone clarify.




Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3
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