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# M03-18

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Intern
Joined: 18 Jan 2018
Posts: 38
Location: India
Concentration: Finance, Marketing
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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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Joined: 02 Sep 2009
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09 Mar 2019, 07:10
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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Joined: 18 Jul 2018
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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??

THANKS
Math Expert
Joined: 02 Sep 2009
Posts: 58465

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01 May 2019, 04:47
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??

THANKS

Hope it helps.
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01 Jul 2019, 22:46
Correct me if i'm wrong but this is simply the difference of squares?

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

Therefore there are 2 distinct roots.

I had dyslexia during the GMATClub test in which this question came up and I thought for some reason it was asking for that formula b^2-root 4ac... forget the name of it!
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Joined: 06 May 2019
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25 Jul 2019, 20:06
The statement becomes

(X^2 -1)^2=0

X=+-1

Thus, we have two values
Re: M03-18   [#permalink] 25 Jul 2019, 20:06

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# M03-18

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