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If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?

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Joined: 02 Sep 2009
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If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?  [#permalink]

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29 Apr 2016, 01:34
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43% (02:15) correct 57% (02:32) wrong based on 469 sessions

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If $$x^2−5x+6=2−|x−1|$$, what is the product of all possible values of x?

A. 1
B. 3
C. 4
D. 5
E. 15

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Joined: 02 Aug 2009
Posts: 7037
Re: If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?  [#permalink]

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29 Apr 2016, 09:06
3
2
Bunuel wrote:
If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?

A. 1
B. 3
C. 4
D. 5
E. 15

Hi,
A very good Q, which tells us to be very careful while dealing with MODULUS
Lets solve it by critical method to avoid any ERROR..

One critical point at x=1.. so TWO ranges

1) x<1..
$$x^2-5x+6=2-{-(x-1)}$$..
$$x^2-5x+6=2+x-1$$..
$$=> x^2-6x+5=0$$ OR$$x^2-5x-x+5=0$$ or $$(x-5)(x-1)=0$$
x=5 or x=1..
BUT since we are looking at the range x<1, BOTH 1 and 5 are not valid..

2) x>=1
$$x^2-5x+6=2-(x-1)$$..
$$x^2-5x+6=2-x+1$$..
$$=> x^2-4x+3=0$$ OR $$x^2-3x-x+3=0$$ or$$(x-3)(x-1)=0$$
x=3 or x=1..
As we are looking at the range x>=1, BOTH 3 and 1 are VALID solution..

so product =1*3=3
B
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?  [#permalink]

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29 Apr 2016, 08:39
1
I solved it by isolating the modulus from the rest of the equation and solving two different quadratic equations:
x-1 = xˆ2-5x+4 -> xˆ2-6x+5=0 -> x1 = 5, x2 = 1
x-1 = -xˆ2+5x-4 -> -xˆ2+4x-3=0 -> x1 = -1, x2 = -3

Solution: 3x5 = 15, letter E
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Re: If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?  [#permalink]

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29 Apr 2016, 08:56
2
3
marcelonac wrote:
I solved it by isolating the modulus from the rest of the equation and solving two different quadratic equations:
x-1 = xˆ2-5x+4 -> xˆ2-6x+5=0 -> x1 = 5, x2 = 1
x-1 = -xˆ2+5x-4 -> -xˆ2+4x-3=0 -> x1 = -1, x2 = -3

Solution: 3x5 = 15, letter E

Hi,

you have found the different values of x correctly as 5,1,3,1..
BUT which are the valid values- you will have to substitute and check- WHY?
Since you are dealing with MODULUS..

$$x^2−5x+6=2−|x−1|$$..

1)5 --
$$5^2-5*5+6=2-|5-1|$$..
$$25-25+6=2-4$$ or $$6=-4$$.. NO, so 5 is not valid..

2)1--
$$1^2-5*1+6=2-|1-1|$$..
$$2=2$$.. VALID

3) 3 --
$$3^2-5*3+6=2-|3-1|$$..
$$9-15+6 = 2-2$$ or $$0=0$$.. VALID..

so there are two valid solutions 1 and 3..
ans 1*3 = 3
B
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?  [#permalink]

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29 Apr 2016, 12:04
Bunuel wrote:
If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?

A. 1
B. 3
C. 4
D. 5
E. 15

Try substituting the values from the options

$$x^2−5x+6=2−|x−1|$$

Option (A)

$$1^2−5*1+6=2−|1−1|$$

$$1−5+6=2−|0|$$

$$10 not =2$$

Option (B)

$$3^2−5*3+6=2−|3−1|$$

$$9−5*3+6=2−2$$

$$15−15=2−2$$

Hence this is our answer , you don't need to go beyond that, to mark the answer as (B)

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Re: If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?  [#permalink]

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29 May 2016, 12:54
I don't understand !

Here, we find the solutions : (1 or 5) for one case | x - 3| = x - 3
and (1 or 3) for the other case | x - 3| = - (x - 3) = - x + 3

This question teaches us that when we solve an equation with modulus, all the answers have to be double-checked ?
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Re: If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?  [#permalink]

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29 May 2016, 20:33
1
Abhishek009 wrote:
Bunuel wrote:
If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?

A. 1
B. 3
C. 4
D. 5
E. 15

Try substituting the values from the options

$$x^2−5x+6=2−|x−1|$$

Option (A)

$$1^2−5*1+6=2−|1−1|$$

$$1−5+6=2−|0|$$

$$10 not =2$$

Option (B)

$$3^2−5*3+6=2−|3−1|$$

$$9−5*3+6=2−2$$

$$15−15=2−2$$

Hence this is our answer , you don't need to go beyond that, to mark the answer as (B)

Yes, I would solve by substituting from the options too but note that I would not try option (A) at all and after option (B), I will need to try option (D) too.

Here are the reasons:
Option (A) is 1. If it satisfies, I am not able to eliminate any options because the product of the values of x would be the same whether you multiply by 1 or not. So no point trying out 1.

Option (B) needs to be tried out to find out whether 3 can be a value of x.

x^2−5x+6=2−|x−1|
3^2−5*3+6=2−|3−1|
0 = 0

So 3 is a value of x. Can we say answer must be (B) yet? No. You have 15 as an option too. So you must try 5 too. If that can also be the value of x, then the product will be 15 and answer will be (E).

5^2−5*5+6=2−|5−1|
6 is not equal to -2.
So 5 is not a value of x.

Now we can be sure that answer must be (B).
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Re: If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?  [#permalink]

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29 May 2016, 20:50
2
Bunuel wrote:
If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?

A. 1
B. 3
C. 4
D. 5
E. 15

Since modulus is given, it is prudent to solve this equation in two parts.

(1) x-1 > 0 or equal to 0. so x is greater than or equal to 1
Then |x−1| can be written as x-1.

The equation can be written as:

$$x^2$$−5x+6=2−(x−1)
$$x^2$$−5x+6=2−x+1
$$x^2$$−5x+6=3−x
$$x^2$$−4x+3=0
(x-1)(x-3) = 0 ; x = 1 or 3. This is in line with our assumption (that x is greater than or equal to 1)

(2) x-1 < 0. so x is less than 1
Then |x−1| can be written as -(x-1).

The equation can be written as:

$$x^2$$−5x+6=2−[-(x−1)]
$$x^2$$−5x+6=2+x-1
$$x^2$$−5x+6=1+x
$$x^2$$−6x+5=0
(x-1)(x-5) = 0; x = 1 or 5. This is NOT in line with our assumption (that x is less than 1)

Acceptable values: 1 and 3.

Product of these values: 1*3 = 3

B is the answer.
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Re: If x^2−5x+6=2−|x−1|, what is the product of all possible values of x?  [#permalink]

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31 Aug 2018, 16:03
case 1 x>=1
x^2-5x+6 = 2-x+1
x = 1,3(acceptable)
case 2 x<1
x^2-5x+6 = 2+x-1
x = 5,1 but x<1 so rejected
Re: If x^2−5x+6=2−|x−1|, what is the product of all possible values of x? &nbs [#permalink] 31 Aug 2018, 16:03
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