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09 Jun 2015, 07:32



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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.
Answer: D 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(x1) = 2 => x = 2 or x1 = 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 31 = 2
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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.
Answer: D 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(x1) = 2 => x = 2 or x1 = 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 31 = 2Substitute x=3 into x^2  2 = x. Does it hold?
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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^22>=0. Then I solve the two cases:
1. x^22 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



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03 Mar 2016, 04:03
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.
Answer: D 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(x1) = 2 => x = 2 or x1 = 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 31 = 2Hi 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^22>=0. Then I solve the two cases: 1. x^22 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



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23 Jun 2016, 07:57
I think this is a highquality question.



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12 Nov 2016, 07:24
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



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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



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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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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



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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



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28 Jan 2017, 21:44
hi Bunuel why are are not considering negative root here ?



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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.



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01 Jun 2017, 22:51



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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 = 21 = 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.
Answer: D 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







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