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Re M0318 [#permalink]
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16 Sep 2014, 00:20



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Re: M0318 [#permalink]
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22 Nov 2014, 21:58
Value of x cant be negative as it equals to the even (4th) root of some expression.
So ,according to me x=1,therefore only one root.
Bunuel please clarify this.



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Re: M0318 [#permalink]
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03 Apr 2016, 08:27
Bunuel wrote: Official Solution:
How many distinct roots does the equation \(x^4  2x^2 + 1=0\) have?
A. 0 B. 1 C. 2 D. 3 E. 4
\(x^4  2x^2 +1=0\); \((x^21)^2=0\); \(x^21=0\); \((x1)(x+1)=0\). Either \(x=1\) or \(x=1\). So, the given equation has two distinct roots.
Answer: C Bunel, Could you pls explain the steps to get to the highlighted step? Thanks, A



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03 Apr 2016, 09:27



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Bunuel wrote: Official Solution:
How many distinct roots does the equation \(x^4  2x^2 + 1=0\) have?
A. 0 B. 1 C. 2 D. 3 E. 4
\(x^4  2x^2 +1=0\); \((x^21)^2=0\); [b]\(x^21=0\); \((x1)(x+1)=0\). Either \(x=1\) or \(x=1\). So, the given equation has two distinct roots.
Answer: C can you please explain how you eliminated the square in the highlighted step. thanks



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13 Dec 2016, 01:56



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Re: M0318 [#permalink]
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01 Apr 2017, 08:54
Bunuel, though this may be a simple question, I always struggle to spot and close an open quadratic to a closed one, as has been done in this question. Is there any simple way to do so?



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02 Apr 2017, 05:12



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Re: M0318 [#permalink]
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19 Jun 2017, 21:53
In the given equation, substitute y = x^2. The equation will reduce to a quadratic equation and the roots of y will 1.
Therefore, x^2 = 1 which means, x can be +1 or 1.



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Re: M0318 [#permalink]
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19 Jun 2017, 22:26
Equation: x^4  2x^2 + 1=0 (x^2  1)^2 = 0 (x^2  1) = 0 x^2 = 1 Hence, x = +/ 1 So, this Equation will have two distinct roots Hence, Answer is C
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Re: M0318 [#permalink]
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09 Sep 2017, 04:02
Bunuel, I rejected 1 because (x2−1)2=0(x2−1)2=0. Could you please provide few links to get a firm grip on such questions. I have seen one solution before in which one value was eliminated because it was raised to the power of 4.
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Re M0318 [#permalink]
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22 Apr 2018, 07:23
I factored x^2: x^2 (x^2  1) = 0 Then I get three solutions: x^2=0, i.e. x=0 x^2=1, i.e. x=1 and x=1
Is it in general possible to factor in that way and, if so, where's the flaw?



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