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Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]
13 Feb 2013, 04:28

1

This post received KUDOS

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 (x-1) & 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. _________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]
13 Feb 2013, 08:14

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 _________________

Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]
13 Feb 2013, 09:13

8

This post received KUDOS

Expert's post

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This post was BOOKMARKED

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| = -(x-1)=1-x, so in this case we'll have x^2 - 4x + 6 = 3-(1-x) --> x^2-5x+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| = x-1, so in this case we'll have x^2 - 4x + 6 = 3-(x-1) --> x^2-3x+2=0 --> x=1 or x=2.

Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]
14 Feb 2013, 10:44

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| = -(x-1)=1-x, so in this case we'll have x^2 - 4x + 6 = 3-(1-x) --> x^2-5x+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| = x-1, so in this case we'll have x^2 - 4x + 6 = 3-(x-1) --> x^2-3x+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?

Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]
15 Feb 2013, 02:43

Expert's post

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| = -(x-1)=1-x, so in this case we'll have x^2 - 4x + 6 = 3-(1-x) --> x^2-5x+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| = x-1, so in this case we'll have x^2 - 4x + 6 = 3-(x-1) --> x^2-3x+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

Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]
18 Feb 2013, 03:27

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| = -(x-1)=1-x, so in this case we'll have x^2 - 4x + 6 = 3-(1-x) --> x^2-5x+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| = x-1, so in this case we'll have x^2 - 4x + 6 = 3-(x-1) --> x^2-3x+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

Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]
18 Feb 2013, 03:35

Expert's post

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| = -(x-1)=1-x, so in this case we'll have x^2 - 4x + 6 = 3-(1-x) --> x^2-5x+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| = x-1, so in this case we'll have x^2 - 4x + 6 = 3-(x-1) --> x^2-3x+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).

Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]
25 Apr 2013, 06:23

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 <?

x+2=0 --> x=-2; OR (x+2)^2-1=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|=-(x-2). 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|=(x-2). 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.

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