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# M31-19

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Math Expert
Joined: 02 Sep 2009
Posts: 53063

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09 Jun 2015, 07:32
00:00

Difficulty:

85% (hard)

Question Stats:

42% (00:59) correct 58% (01:08) wrong based on 72 sessions

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What is the range of all the roots of $$|x^2 - 2| = x$$?

A. 4
B. 3
C. 2
D. 1
E. 0

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 53063

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09 Jun 2015, 07:32
1
2
Official Solution:

What is the range of all the roots of $$|x^2 - 2| = x$$?

A. 4
B. 3
C. 2
D. 1
E. 0

First of all notice that since $$x$$ is equal to an absolute value of some number ($$|x^2 - 2|$$), then $$x$$ cannot be negative.

Next, $$|x^2 - 2| = x$$ means that either $$x^2 - 2 = x$$ or $$-(x^2 - 2) = x$$.

First equation gives $$x = -1$$ or $$x = 2$$. Since $$x$$ cannot be negative, we are left with only $$x = 2$$.

Second equation gives $$x = -2$$ or $$x = 1$$. Again, since $$x$$ cannot be negative, we are left with only $$x = 1$$.

The range = {largest} - {smallest} = 2 - 1 = 1.

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Joined: 18 Sep 2014
Posts: 227

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30 Jan 2016, 01:49
Bunuel wrote:
Official Solution:

What is the range of all the roots of $$|x^2 - 2| = x$$?

A. 4
B. 3
C. 2
D. 1
E. 0

First of all notice that since $$x$$ is equal to an absolute value of some number ($$|x^2 - 2|$$), then $$x$$ cannot be negative.

Next, $$|x^2 - 2| = x$$ means that either $$x^2 - 2 = x$$ or $$-(x^2 - 2) = x$$.

First equation gives $$x = -1$$ or $$x = 2$$. Since $$x$$ cannot be negative, we are left with only $$x = 2$$.

Second equation gives $$x = -2$$ or $$x = 1$$. Again, since $$x$$ cannot be negative, we are left with only $$x = 1$$.

The range = {largest} - {smallest} = 2 - 1 = 1.

Hi,
Could you kindly explain the below in red:
First of all notice that since x is equal to an absolute value of some number (|x 2 −2| ), then x cannot be negative.

Next, |x 2 −2|=x means that either x 2 −2=x or −(x 2 −2)=x .

First equation gives x=−1 or x=2 x^2 - 2 = x => x^2 - x = 2 => x(x-1) = 2 => x = 2 or x-1 = 2 (x=3) . Since x cannot be negative, we are left with only x=2 (should be 3) .

Second equation gives x=−2 or x=1 . Again, since x cannot be negative, we are left with only x=1 .

The range = {largest} - {smallest} = 2 - 1 = 1. IMO it should be 3-1 = 2
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Math Expert
Joined: 02 Sep 2009
Posts: 53063

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31 Jan 2016, 04:26
FightToSurvive wrote:
Bunuel wrote:
Official Solution:

What is the range of all the roots of $$|x^2 - 2| = x$$?

A. 4
B. 3
C. 2
D. 1
E. 0

First of all notice that since $$x$$ is equal to an absolute value of some number ($$|x^2 - 2|$$), then $$x$$ cannot be negative.

Next, $$|x^2 - 2| = x$$ means that either $$x^2 - 2 = x$$ or $$-(x^2 - 2) = x$$.

First equation gives $$x = -1$$ or $$x = 2$$. Since $$x$$ cannot be negative, we are left with only $$x = 2$$.

Second equation gives $$x = -2$$ or $$x = 1$$. Again, since $$x$$ cannot be negative, we are left with only $$x = 1$$.

The range = {largest} - {smallest} = 2 - 1 = 1.

Hi,
Could you kindly explain the below in red:
First of all notice that since x is equal to an absolute value of some number (|x 2 −2| ), then x cannot be negative.

Next, |x 2 −2|=x means that either x 2 −2=x or −(x 2 −2)=x .

First equation gives x=−1 or x=2 x^2 - 2 = x => x^2 - x = 2 => x(x-1) = 2 => x = 2 or x-1 = 2 (x=3) . Since x cannot be negative, we are left with only x=2 (should be 3) .

Second equation gives x=−2 or x=1 . Again, since x cannot be negative, we are left with only x=1 .

The range = {largest} - {smallest} = 2 - 1 = 1. IMO it should be 3-1 = 2

Substitute x=3 into x^2 - 2 = x. Does it hold?
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Joined: 03 Oct 2014
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03 Mar 2016, 04:01
Hi Bunuel,

is it correct to solve like usual the abs value function for both the cases and then put alltogether in a system?
In this case, I have that everything is above the root should be >= 0, |x^2-2|>=0. Then I solve the two cases:

1. x^2-2 for x^2>=2 I obtain: x<=-1 x>=2 (only x>=2 is acceptable
2.- x^2+2 for x^2<=2 I obtain: x<=-1 -2<x<=1 acceptable range

Now if I pu all togheter to have a solution that is >=0: x<-2 and 1<x<2 so I take only the second range?
Correct me on my reasoning .

Thanks
Intern
Joined: 03 Oct 2014
Posts: 3

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03 Mar 2016, 04:03
1
FightToSurvive wrote:
Bunuel wrote:
Official Solution:

What is the range of all the roots of $$|x^2 - 2| = x$$?

A. 4
B. 3
C. 2
D. 1
E. 0

First of all notice that since $$x$$ is equal to an absolute value of some number ($$|x^2 - 2|$$), then $$x$$ cannot be negative.

Next, $$|x^2 - 2| = x$$ means that either $$x^2 - 2 = x$$ or $$-(x^2 - 2) = x$$.

First equation gives $$x = -1$$ or $$x = 2$$. Since $$x$$ cannot be negative, we are left with only $$x = 2$$.

Second equation gives $$x = -2$$ or $$x = 1$$. Again, since $$x$$ cannot be negative, we are left with only $$x = 1$$.

The range = {largest} - {smallest} = 2 - 1 = 1.

Hi,
Could you kindly explain the below in red:
First of all notice that since x is equal to an absolute value of some number (|x 2 −2| ), then x cannot be negative.

Next, |x 2 −2|=x means that either x 2 −2=x or −(x 2 −2)=x .

First equation gives x=−1 or x=2 x^2 - 2 = x => x^2 - x = 2 => x(x-1) = 2 => x = 2 or x-1 = 2 (x=3) . Since x cannot be negative, we are left with only x=2 (should be 3) .

Second equation gives x=−2 or x=1 . Again, since x cannot be negative, we are left with only x=1 .

The range = {largest} - {smallest} = 2 - 1 = 1. IMO it should be 3-1 = 2

Hi Bunuel,

is it correct to solve like usual the abs value function for both the cases and then put alltogether in a system?
In this case, I have that everything is above the root should be >= 0, |x^2-2|>=0. Then I solve the two cases:

1. x^2-2 for x^2>=2 I obtain: x<=-1 x>=2 (only x>=2 is acceptable
2.- x^2+2 for x^2<=2 I obtain: x<=-1 -2<x<=1 acceptable range

Now if I pu all togheter to have a solution that is >=0: x<-2 and 1<x<2 so I take only the second range?
Correct me on my reasoning .

Thanks
Current Student
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Posts: 4
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GRE 1: Q160 V160
GPA: 3.98
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23 Jun 2016, 07:57
I think this is a high-quality question.
Intern
Joined: 31 Oct 2016
Posts: 2

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12 Nov 2016, 07:24
1
I believe there is a small correction needed in the explanation:

Scenario I: For x^2 - 2 > 0 i.e. x > Sqrt(2) then x=2, x=-1, the only acceptable solution for this scenario is x=2
Scenario II: For x^2 - 2 < 0 i.e. - Sqrt(2) < x < Sqrt(2), then x=-2, x=1. the acceptable solution for this scenario is x=1

Therfore the range xmax - xmin = 2 - 1 = 1

Thanks for all the great work
Math Expert
Joined: 02 Sep 2009
Posts: 53063

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12 Nov 2016, 07:31
jcuchet wrote:
I believe there is a small correction needed in the explanation:

Scenario I: For x^2 - 2 > 0 i.e. x > Sqrt(2) then x=2, x=-1, the only acceptable solution for this scenario is x=2
Scenario II: For x^2 - 2 < 0 i.e. - Sqrt(2) < x < Sqrt(2), then x=-2, x=1. the acceptable solution for this scenario is x=1

Therfore the range xmax - xmin = 2 - 1 = 1

Thanks for all the great work

_______________
What correction?
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Joined: 31 Oct 2016
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12 Nov 2016, 07:36
Bunuel wrote:
jcuchet wrote:
I believe there is a small correction needed in the explanation:

Scenario I: For x^2 - 2 > 0 i.e. x > Sqrt(2) then x=2, x=-1, the only acceptable solution for this scenario is x=2
Scenario II: For x^2 - 2 < 0 i.e. - Sqrt(2) < x < Sqrt(2), then x=-2, x=1. the acceptable solution for this scenario is x=1

Therfore the range xmax - xmin = 2 - 1 = 1

Thanks for all the great work

_______________
What correction?

_______________
Scenario II is for - Sqrt(2) < x < Sqrt(2) not just for x < 0. Therefore under scenario II a negative solution is allowable between -Sqrt(2) and 0
Math Expert
Joined: 02 Sep 2009
Posts: 53063

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12 Nov 2016, 07:41
jcuchet wrote:
Bunuel wrote:
jcuchet wrote:
I believe there is a small correction needed in the explanation:

Scenario I: For x^2 - 2 > 0 i.e. x > Sqrt(2) then x=2, x=-1, the only acceptable solution for this scenario is x=2
Scenario II: For x^2 - 2 < 0 i.e. - Sqrt(2) < x < Sqrt(2), then x=-2, x=1. the acceptable solution for this scenario is x=1

Therfore the range xmax - xmin = 2 - 1 = 1

Thanks for all the great work

_______________
What correction?

_______________
Scenario II is for - Sqrt(2) < x < Sqrt(2) not just for x < 0. Therefore under scenario II a negative solution is allowable between -Sqrt(2) and 0

No, that's not correct. x cannot be negative for any scenario. This is explained in the first sentence of the solution:

First of all notice that since $$x$$ is equal to an absolute value of some number ($$|x^2 - 2|$$), then $$x$$ cannot be negative.
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Joined: 10 Dec 2015
Posts: 12
GMAT 1: 590 Q36 V35

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26 Nov 2016, 18:46
Hi, could you please explain how the first equation gives x=−1 or x=2? I got x=2 and x=3.
Thank you very much
Math Expert
Joined: 02 Sep 2009
Posts: 53063

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27 Nov 2016, 00:34
nelliegu wrote:
Hi, could you please explain how the first equation gives x=−1 or x=2? I got x=2 and x=3.
Thank you very much

x^2 - x - 2 = 0
(x - 2) (x + 1) = 0
x = 2 or x = -1.

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Intern
Joined: 10 Dec 2015
Posts: 12
GMAT 1: 590 Q36 V35

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27 Nov 2016, 08:00
Bunuel wrote:
nelliegu wrote:
Hi, could you please explain how the first equation gives x=−1 or x=2? I got x=2 and x=3.
Thank you very much

x^2 - x - 2 = 0
(x - 2) (x + 1) = 0
x = 2 or x = -1.

Of course!! thank you very much Bunuel
Manager
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28 Jan 2017, 21:44
hi Bunuel
why are are not considering negative root here ?
Math Expert
Joined: 02 Sep 2009
Posts: 53063

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29 Jan 2017, 02:54
gupta87 wrote:
hi Bunuel
why are are not considering negative root here ?

This is explained in the first sentence of the solution:

First of all notice that since $$x$$ is equal to an absolute value of some number ($$|x^2 - 2|$$), then $$x$$ cannot be negative.
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Joined: 26 Nov 2012
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01 Jun 2017, 22:28
Bunuel wrote:
gupta87 wrote:
hi Bunuel
why are are not considering negative root here ?

This is explained in the first sentence of the solution:

First of all notice that since $$x$$ is equal to an absolute value of some number ($$|x^2 - 2|$$), then $$x$$ cannot be negative.

Bunuel, can you please share all conditions, under which cases $$x$$ is equal to an absolute value of some number is positive or negative. I didn't find the above highlighted point in our math book.
Math Expert
Joined: 02 Sep 2009
Posts: 53063

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01 Jun 2017, 22:51
msk0657 wrote:
Bunuel wrote:
gupta87 wrote:
hi Bunuel
why are are not considering negative root here ?

This is explained in the first sentence of the solution:

First of all notice that since $$x$$ is equal to an absolute value of some number ($$|x^2 - 2|$$), then $$x$$ cannot be negative.

Bunuel, can you please share all conditions, under which cases $$x$$ is equal to an absolute value of some number is positive or negative. I didn't find the above highlighted point in our math book.

x = |some expression| in any case means that x cannot be negative, it can be 0 or positive.
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02 Jun 2017, 07:40
the question can be solved in another way too..
square both sides of the equation
the resulting equation is
x^4 - 5x^2 +4 = 0

solving for x^2 gives 4 and 1 as roots
implies x= +/- 2 or +/- 1

range = 2-1 = 1

ans D
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03 Jun 2017, 02:14
Bunuel wrote:
Official Solution:

What is the range of all the roots of $$|x^2 - 2| = x$$?

A. 4
B. 3
C. 2
D. 1
E. 0

First of all notice that since $$x$$ is equal to an absolute value of some number ($$|x^2 - 2|$$), then $$x$$ cannot be negative.

Next, $$|x^2 - 2| = x$$ means that either $$x^2 - 2 = x$$ or $$-(x^2 - 2) = x$$.

First equation gives $$x = -1$$ or $$x = 2$$. Since $$x$$ cannot be negative, we are left with only $$x = 2$$.

Second equation gives $$x = -2$$ or $$x = 1$$. Again, since $$x$$ cannot be negative, we are left with only $$x = 1$$.

The range = {largest} - {smallest} = 2 - 1 = 1.

Bunuel,

I used the highlighted part above as a base for square both sides and arrived to same solutions presented above. Is my basis valid or not?

Thakns
Re: M31-19   [#permalink] 03 Jun 2017, 02:14

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# M31-19

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