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M03-18

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M03-18  [#permalink]

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New post 15 Sep 2014, 23:20
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New post 15 Sep 2014, 23:20
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


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

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

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

\((x-1)(x+1)=0\). Either \(x=1\) or \(x=-1\). So, the given equation has two distinct roots.


Answer: C
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Re: M03-18  [#permalink]

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New post 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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Re: M03-18  [#permalink]

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New post 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


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

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

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

\((x-1)(x+1)=0\). 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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Re: M03-18  [#permalink]

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New post 03 Apr 2016, 08:27
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


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

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

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

\((x-1)(x+1)=0\). 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: \(a^2-2ab+b^2=(a-b)^2\)
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M03-18  [#permalink]

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New post 12 Dec 2016, 18:45
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


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

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

[b]\(x^2-1=0\);

\((x-1)(x+1)=0\). 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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Re: M03-18  [#permalink]

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New post 13 Dec 2016, 00:56
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


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

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

[b]\(x^2-1=0\);

\((x-1)(x+1)=0\). 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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Re: M03-18  [#permalink]

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New post 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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New post 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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Re: M03-18  [#permalink]

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New post 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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Re: M03-18  [#permalink]

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New post 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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New post 09 Sep 2017, 03:30
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 Megathread

Hope it helps.
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New post 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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New post 22 Apr 2018, 20:35
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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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New post 15 Sep 2018, 03:27
Hi Bunuel

I 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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New post 15 Sep 2018, 03:40
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ENEM wrote:
Hi Bunuel

I 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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New post 15 Sep 2018, 04:25
Bunuel wrote:
ENEM wrote:
Hi Bunuel

I 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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New post 15 Sep 2018, 04:28
1
ENEM wrote:
Bunuel wrote:
ENEM wrote:
Hi Bunuel

I 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/m03-183603.html#p2050204
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Re: M03-18 &nbs [#permalink] 15 Sep 2018, 04:28
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