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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
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
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
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
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.



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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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20 Jun 2018, 07:13
JeffTargetTestPrep wrote: 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 generis hello, every time i think of modulus/absolute values, i think of you do have any idea why the product of all solutions is 6 and not 12 ? we have case 1.) x = 2 or x = 1 and case 2.) x = 3 or x = 2 so product of all the solutions (2)(1)(3)(2) thank you very much



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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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20 Jun 2018, 14:41
EMPOWERgmatRichC wrote: 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 EMPOWERgmatRichC hello there selfmade GMAT assassin
you know i dont understand the below explanation...have some questions
Why x =0 ? "Looking at the answer choices, they're all within a relatively small distance from 0" answer choices range from 2 to 12 how can they be close to zero ? and how does the fact that they are they're all within a relatively small distance from 0 to solve this problem ? \(X^2+4X+7=X+2+3\) 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"... from here X^2+4X+7=X+2+3 from here opening modulus gives negative and positive cases :negative \(X+2+3\)= \((x+2)+3\) = \(x2+3\) = \(x1\) hence \(X^2+4X+7=x1\) > \(X^2+5X+8=0\) now what ? to find discriminant ? \(b^24ac\) ? ok \(25  4 (8) =  7\) now i got stuck :positive \(X+2+3\) > \(x+2+3\) so i get \(X^2+4X+7=x+2+3\) > \(X^2+4X+7=x+5\) > \(X^2+3X+2=0\) now ? X = 0 where from did you get zero ? (0+2)(0+2) = 0+2? how did you get this ? and why you pluged in zero ? (2)(2) = 2? NO, 0 is NOT a solution could you please take time to explain thank you



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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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20 Jun 2018, 19:04
dave13 wrote: JeffTargetTestPrep wrote: 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 generis hello, every time i think of modulus/absolute values, i think of you do have any idea why the product of all solutions is 6 and not 12 ? we have case 1.) x = 2 or x = 1 and case 2.) x = 3 or x = 2 so product of all the solutions (2)(1)(3)(2) thank you very much Hi dave13Quote: every time i think of modulus/absolute values, i think of you Oh, how my cohorts could (and likely will) have fun with that one! I'm having fun with it. You have a great sense of humor. Thanks for the laugh. Quote: why the product of all solutions is []6 and not 12 ? This question was asked by your predecessors Nived and khan0210 ( khan0210, I think you understand the issue. See below, and please see link below.* Sorry you all had to wait so long.) dave13The answer is \(6\), and not \(12\), because one of the \((x = 2)\) solutions is not valid. That is, there are only three valid solutions to this equation. Those three solutions are 3, 2, and 1 The problem comes when we consider the case that \(x + 2\) is positive. When \((x + 2)\) is positive (one of two "original conditions"): \(x + 2 > 0\) \(x > 2\)and \(x+2 = (x + 2)\) Solving for when \(x + 2 >0\) \(x^2 + 4x + 7 = (x + 2) + 3\) \(x^2 + 4x + 7 = x + 5\) \(x^2 + 3x + 2 = 0\) \((x + 2)(x + 1) = 0\) \(x = 2\) or \(x = 1\) x cannot = 2x cannot be both greater than 2 and equal to 2.\((x = 2)\) does not satisfy the original condition that \(x > 2\) (from \(x + 2 > 0\)) The solution \((x = 2)\) falls outside the range we are investigating, and hence is not valid. The only valid solution in this range is \((x = 1)\) So there are three solutions: 3, 2, and 1. Their product is 6 This material is challenging. Here is a GMAT Club post on how to treat Absolute ValueI hope that helps! khan0210 , I read your reasoning, and it appears to me that you understand the source of the problem. I make that conclusion based on what you wrote here:
Quote: checking: 2+2>0 0>0 {null so doesn't apply}
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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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Updated on: 21 Jun 2018, 02:27
Solution Given:• We are given an equation : \(x^2 + 4x + 7 = x + 2 + 3\)
To find:• We need to find the product of all the values of x that satisfy the equation \(x^2 + 4x + 7 = x + 2 + 3\) Approach and Working: Observe that we have x+2 in our equation and its values will be : • (x+2) for x<2 • (x+2) for x>= 2 Thus, we need to consider the equation for x<2 and for x>=2 differently as the equation will change for both the ranges. For x<2 :• The equation will be\(x^2 + 4x + 7 = x  2 + 3\)
o \(x^2+5x+6=0\) o x= 2,3 For x>=2 :• The equation will be \(x^2 + 4x + 7 = x + 2 + 3\)
o \(x^2+3x+2=0\) o x= 1, 2 Thus, the product of all the values of x is: 1*2*3= 6 Hence, the correct answer is option A. 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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21 Jun 2018, 02:23
EgmatQuantExpert wrote: Solution Given:• We are given an equation : \(x^2 + 4x + 7 = x + 2 + 3\)
To find:• We need to find the product of all the values of x that satisfy the equation \(x^2 + 4x + 7 = x + 2 + 3\) Approach and Working: Observe that we have x+2 in our equation and its values will be : • (x+2) for x<2 • (x+2) for x>= 2 Thus, we need to consider the equation for x<2 and for x>=2 differently as the equation will change for both the ranges. For x<2 :• The equation will be\(x^2 + 4x + 7 = x  2 + 3\)
For x>=2 :• The equation will be \(x^2 + 4x + 7 = x + 2 + 3\)
o \(x^2+3x+2=0\) o x= 1, 2 Thus, the product of all the values of x is: 1*2*2*3= 12 Hence, the correct answer is option E. Answer: EHey EgmatQuantExpert are you sure its E ? the correct answer is 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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Re: What is the product of all the solutions of x^2 + 4x + 7 = x + 2 + 3 [#permalink]
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21 Jun 2018, 03:53
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 Some testtakers might find it easiest to proceed as follows: 1. Solve the equation with the signs unchanged 2. Solve the equation with the signs changed in the absolute value In each case, plug any possible solutions back into the original equation to confirm that they are valid. \(x^2 + 4x + 7 = x + 2 + 3\) \(x^2 + 4x + 4 = x + 2\) Here, a solution will be valid as along as the left side is NONNEGATIVE. Case 1: signs unchanged\(x^2 + 4x + 4 = x + 2\) \(x^2 + 3x + 2 = 0\) \((x+2)(x+1) = 0\) Possible solutions: x=2, x=1. If we plug \(x=2\) into \(x^2 + 4x + 4\), we get: \((2)^2 + (4)(2) + 4 = 4  8 + 4 = 0\). Since the result is nonnegative, x=2 is a valid solution. If we plug \(x=1\) into \(x^2 + 4x + 4\), we get: \((1)^2 + (4)(1) + 4 = 1  4 + 4 = 1\). Since the result is nonnegative, x=1 is a valid solution. Case 2: signs changed in the absolute value\(x^2 + 4x + 4 = x  2\) \(x^2 + 5x + 6 = 0\) \((x+3)(x+2) = 0\) Possible solutions: x=3, x=2. Since x=2 was a valid solution in Case 1, only x=3 must be tested. If we plug \(x=3\) into \(x^2 + 4x + 4\), we get: \((3)^2 + (4)(3) + 4 = 9  12 + 4 = 1\). Since the result is nonnegative, x=3 is a valid solution. Product of the solutions = (2)(1)(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 Jun 2018, 11:05
Hi all,
I approached the question slightly different in the way I tried to solve.
Firstly, when the righthand side is positive I set my equation to (x+2)(x+2)=(x+2), thus I divided both sides by (x+2) and was left with x+2=1, which gives me x=1. However, when the RHS is negative I get (x+2)(x+2)=  (x+2); thus when I divide by x+2, I get x+2=1, which yields x=3. So, I am missing x=2.
What is wrong with my approach here? Why can I not divide out (x+2)?




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