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# M03-18

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Math Expert
Joined: 02 Sep 2009
Posts: 55804

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16 Sep 2014, 00:20
2
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Difficulty:

15% (low)

Question Stats:

71% (00:54) correct 29% (00:52) wrong based on 233 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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Math Expert
Joined: 02 Sep 2009
Posts: 55804

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16 Sep 2014, 00:20
1
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.

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Joined: 06 Apr 2014
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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.

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Joined: 02 Sep 2009
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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.

That's not correct. Plug -1 into the equation, does it hold true?
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Schools: ISB '18, SPJ GMBA '17
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03 Apr 2016, 08:27
1
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.

Bunel,

Could you pls explain the steps to get to the highlighted step?

Thanks,
A
Math Expert
Joined: 02 Sep 2009
Posts: 55804

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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

$$x^4 - 2x^2 +1=0$$;

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

$$x^2-1=0$$;

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

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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Joined: 18 May 2016
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12 Dec 2016, 19:45
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.

can you please explain how you eliminated the square in the highlighted step. thanks
Math Expert
Joined: 02 Sep 2009
Posts: 55804

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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

$$x^4 - 2x^2 +1=0$$;

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

[b]$$x^2-1=0$$;

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

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?
Math Expert
Joined: 02 Sep 2009
Posts: 55804

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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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Joined: 04 Apr 2017
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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

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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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When nothing seem to help, I would go and look at a Stonecutter hammering away at his rock perhaps a hundred time without as much as a crack showing in it.
Yet at the hundred and first blow it would split in two.
And I knew it was not that blow that did it, But all that had gone Before
.
Math Expert
Joined: 02 Sep 2009
Posts: 55804

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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 Megathread

Hope it helps.
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Joined: 07 Apr 2018
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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?
Math Expert
Joined: 02 Sep 2009
Posts: 55804

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22 Apr 2018, 21:35
1
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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Joined: 16 Nov 2016
Posts: 274

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15 Sep 2018, 04:27
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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Math Expert
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15 Sep 2018, 04:40
1
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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15 Sep 2018, 05:25
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??
_________________
If you find my post useful, please give me a kudos.

Thank you.
Regards,
ENEM

If you wish to spend wisely on your gmat prep material, check my post titled: How to Spend Money On GMAT Material Wisely, link: https://gmatclub.com/forum/how-to-buy-gmat-material-wisely-tag-free-gmat-resources-236174.html

Simple and handy template for CR: https://gmatclub.com/forum/simple-and-handy-template-for-cr-242255.html

simple template for more vs greater and fewer vs less: https://gmatclub.com/forum/simple-template-for-more-vs-greater-and-fewer-vs-less-242216.html
Math Expert
Joined: 02 Sep 2009
Posts: 55804

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15 Sep 2018, 05:28
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   [#permalink] 15 Sep 2018, 05:28

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# M03-18

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