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How many distinct roots does the equation \(x^4  2x^2 + 1=0\) have? A. 0 B. 1 C. 2 D. 3 E. 4
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16 Sep 2014, 00:20
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
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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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23 Nov 2014, 05:58
gmattaker10 wrote: 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. That's not correct. Plug 1 into the equation, does it hold true?
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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
ArunpriyanJ wrote: 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 It's a simple algebraic property: \(a^22ab+b^2=(ab)^2\)
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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
bjklue wrote: 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 This is basics: number^2 = 0 > number = 0.
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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
OreoShake wrote: 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? I guess the only way is through practice...
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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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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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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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09 Sep 2017, 04:30
Prashant10692 wrote: 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. 7. Algebra For more check Ultimate GMAT Quantitative MegathreadHope it helps.
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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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22 Apr 2018, 21:35
gmatprep001 wrote: 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? You would be correct if we had x^4  x^2 = 0 but we have \(x^4  2x^2 + 1=0\).
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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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15 Sep 2018, 04:40
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.
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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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15 Sep 2018, 05:28
ENEM wrote: 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?? Yes, that's not correct you got the correct answer by fluke. Check my response here: https://gmatclub.com/forum/m03183603.html#p2050204
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