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M03-18

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Bunuel
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M03-18 [#permalink] New post   16 Sep 2014, 00:20
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74% (00:53) correct 26% (00:57) wrong based on 906 sessions
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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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C
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Bunuel
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Re M03-18 [#permalink] New post   16 Sep 2014, 00:20
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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


We can apply the identity \(a^2 - 2ab + b^2 = (a - b)^2\) to the equation:

\(x^4 - 2x^2 + 1 = 0\);

\((x^2-1)^2=0\);

\(x^2-1 = 0\);

\(x^2=1\).

Thus, the roots are \(x=1\) and \(x=-1\). Therefore, the given equation has two distinct roots.


Answer: C
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Bunuel
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Re: M03-18 [#permalink] New post   09 Sep 2017, 04:30
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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 Megathread

Hope it helps.
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Re: M03-18 [#permalink] New post   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: M03-18 [#permalink] New post   09 Mar 2019, 05:21
Bunuel in this question, I took x = (x^2) and hence calculated the roots for (x^2)^2-2(x^2)+1 by the formula root(B^2-4ac) which turned out to be 0 and hence, I marked choice B. Am I missing something?
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Bunuel
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Re: M03-18 [#permalink] New post   09 Mar 2019, 07:10
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aalakshaya wrote:
Bunuel in this question, I took x = (x^2) and hence calculated the roots for (x^2)^2-2(x^2)+1 by the formula root(B^2-4ac) which turned out to be 0 and hence, I marked choice B. Am I missing something?



\((x^2)^2-2(x^2)+1 =0\)

\(x^2=\frac{2+\sqrt{4-4}}{2}\) --> \(x^2=1\) --> \(x=1\) or \(x = -1\);

\(x^2=\frac{2-\sqrt{4-4}}{2}\) --> \(x^2=1\) --> \(x=1\) or \(x = -1\).
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Re: M03-18 [#permalink] New post   01 May 2019, 04:36
Bunuel wrote:
aalakshaya wrote:
Bunuel in this question, I took x = (x^2) and hence calculated the roots for (x^2)^2-2(x^2)+1 by the formula root(B^2-4ac) which turned out to be 0 and hence, I marked choice B. Am I missing something?



\((x^2)^2-2(x^2)+1 =0\)

\(x^2=\frac{2+\sqrt{4-4}}{2}\) --> \(x^2=1\) --> \(x=1\) or \(x = -1\);

\(x^2=\frac{2-\sqrt{4-4}}{2}\) --> \(x^2=1\) --> \(x=1\) or \(x = -1\).



Hi Bunuel,

I was wondering how did you find the roots as illustrated above??
Is this some kind of formula??

Would appreciate your help!
THANKS
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Bunuel
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Re: M03-18 [#permalink] New post   01 May 2019, 04:47
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JIAA wrote:
Bunuel wrote:
aalakshaya wrote:
Bunuel in this question, I took x = (x^2) and hence calculated the roots for (x^2)^2-2(x^2)+1 by the formula root(B^2-4ac) which turned out to be 0 and hence, I marked choice B. Am I missing something?



\((x^2)^2-2(x^2)+1 =0\)

\(x^2=\frac{2+\sqrt{4-4}}{2}\) --> \(x^2=1\) --> \(x=1\) or \(x = -1\);

\(x^2=\frac{2-\sqrt{4-4}}{2}\) --> \(x^2=1\) --> \(x=1\) or \(x = -1\).



Hi Bunuel,

I was wondering how did you find the roots as illustrated above??
Is this some kind of formula??

Would appreciate your help!
THANKS


Solving and Factoring Quadratics:
https://www.purplemath.com/modules/solvquad.htm
https://www.purplemath.com/modules/factquad.htm

Hope it helps.
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Re: M03-18 [#permalink] New post   23 Mar 2023, 01:35
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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bhavik29
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Re: M03-18 [#permalink] New post   13 Aug 2023, 20:55
What am I doing wrong?

x^4 - 2x^2 + 1 = 0
x^2(x^2-2) + 1 = 0
x^2-2=0 or x^2 + 1 = 0
x^2= 2 or x^2 = -1
x= square root 2
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Bunuel
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Re: M03-18 [#permalink] New post   14 Aug 2023, 01:33
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bhavik29 wrote:
What am I doing wrong?

x^4 - 2x^2 + 1 = 0
x^2(x^2-2) + 1 = 0
x^2-2=0 or x^2 + 1 = 0
x^2= 2 or x^2 = -1
x= square root 2


Nothin there is correct. I think you'd benefit if you brush up algebra fundamentals.

7. Algebra



For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: M03-18 [#permalink] New post   14 Aug 2023, 14:47
x^4 - 2x^2 + 1 = 0

let u = x^2

u^2 - 2u + 1
(u-1)(u-1)
Sub in x^2 in u
(x^2-1)(x^2-1)
(x-1)(x+1)(x-1)(x+1)
so solutions are x=1 and x=-1
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Re: M03-18 [#permalink]
14 Aug 2023, 14:47
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