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15 Sep 2014, 23:20



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22 Nov 2014, 20:58
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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03 Apr 2016, 07:27
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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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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01 Apr 2017, 07:54
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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19 Jun 2017, 20:53
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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19 Jun 2017, 21:26
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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09 Sep 2017, 03:02
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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22 Apr 2018, 06:23
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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Hi BunuelI did this: \(x^2(x^2  2) = 1\) \(x^2 = 1\) therefore x has two roots 1 and 1 is this approach correct?
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Bunuel wrote: ENEM wrote: Hi BunuelI did this: \(x^2(x^2  2) = 1\) \(x^2 = 1\) therefore x has two roots 1 and 1 is this approach correct? The answer is correct. Not clear though how you identified that x^2 = 1 from x^2(x^2  2) = 1 or how would you get the answer if it were x^2(x^2  2) = 10. Bunuel \(x^2(x^2 − 2)= −1\) \(x^2 = −1\) and\((x^2−2) = −1\) this \(x^2 = −1\) is not possible as square of a number cannot be negative. thus we are left with \(x^2−2 = −1\) adding 2 on both sides we get \(x^2=1\) therefore x has two roots 1 and 1 did I goof up??
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