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m03#18 - NS (Polynomial Equations)

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m03#18 - NS (Polynomial Equations) [#permalink] New post 19 May 2010, 08:31
I couldnt locate dscussion of this question.

How many roots does the equation x^4 - 2x^2 +1=0 have?

a) 0
b) 1
c) 2
d) 3
e) 4

As per the solution given, x = 1 or x = -1.

My question is:
We are asked how many roots are there for this equation not how many different roots. So, total no. of roots should be 4.

Experts please comment.
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Re: m03#18 - NS (Polynomial Equations) [#permalink] New post 19 May 2010, 13:03
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Hi ykaiim,

'Roots' aren't like prime factors; repeating a root isn't meaningful. In prime factorization, 2 is the only distinct prime factor of 2 and 32, but since 2 goes into 2 once and 32 five times, 32 have five prime factors. With roots, however, either a given value of X does correctly balance the equation, or it doesn't; it's a meaningless distinction to say that -1 can 'solve the equation twice.' Thus, since there are exactly two x values that solve the equation, 1 and -1, we say there are exactly two roots, even if as a 4th degree equation each of those roots could be derived two ways.
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Re: m03#18 - NS (Polynomial Equations) [#permalink] New post 30 May 2010, 06:07
ykaiim wrote:
I couldnt locate dscussion of this question.

How many roots does the equation x^4 - 2x^2 +1=0 have?

a) 0
b) 1
c) 2
d) 3
e) 4

As per the solution given, x = 1 or x = -1.

My question is:
We are asked how many roots are there for this equation not how many different roots. So, total no. of roots should be 4.

Experts please comment.

technically, number of the roots of polynomial is max power in polynomial. so that way answer must be E.
i.e. number of roots of given equation is 4. out 4, 2 are same. here we are not concerned whether roots are repeated or not. we are asked to find number of roots. ..please comment. thanks
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Re: m03#18 - NS (Polynomial Equations) [#permalink] New post 13 Sep 2011, 03:39
I tried to solve it as below -

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

b= -2
a=1
c=1

Discriminant = b^2-4ac
= (-2)^2 - 4*1*1
=4-4
=0

As b^2-4ac = 0 therefore this equation should have 1 root but ans is 2 roots.

Can someone please advise what I am doing wrong?
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Re: m03#18 - NS (Polynomial Equations) [#permalink] New post 13 Sep 2011, 03:49
agarwalmanoj2000 wrote:
I tried to solve it as below -

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

b= -2
a=1
c=1

Discriminant = b^2-4ac
= (-2)^2 - 4*1*1
=4-4
=0

As b^2-4ac = 0 therefore this equation should have 1 root but ans is 2 roots.

Can someone please advise what I am doing wrong?


You have applied the Discriminant formula of a quadratic equation for a polynomial of 4th degree.
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Re: m03#18 - NS (Polynomial Equations) [#permalink] New post 13 Sep 2011, 04:57
How many roots does the equation x^4 - 2x^2 +1=0 have?

let x^2=a ,then we have a^2-2a+1=0 or (a-1)^2=0

replace a with x-
(a-1)^2=0
(x^2-1)^2=0 or
((x-1)(x+1))^2=0

x=1 x=-1

answ is 2
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Re: m03#18 - NS (Polynomial Equations) [#permalink] New post 13 Sep 2011, 05:42
Thanks, Fluke !!!

Sorry, I did not understand :? .

Can you please provide some more details?

Thank you in advance.

@lalab
Thanks for your response. I am trying to find problem in my approach.
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Re: m03#18 - NS (Polynomial Equations) [#permalink] New post 13 Sep 2011, 07:46
agarwalmanoj2000 wrote:
Thanks, Fluke !!!

Sorry, I did not understand :? .

Can you please provide some more details?

Thank you in advance.

@lalab
Thanks for your response. I am trying to find problem in my approach.


ax^2+bx+c=0. This is a polynomial of degree 2, where D=b^2-4ac

For a polynomial of degree 4, i.e. maximum power of x, falls in a different category:
ax^4+bx^2+c=0. This is a polynomial of degree 4, for which D = b^2-4ac may not hold true.

The question is asking about:
x^4 - 2x^2 +1=0. The maximum power of x is 4. You just don't know what is the discriminant. The discriminant formula you used is only for quadratic equations(x with highest power of 2).

For a quadratic equation, your approach would be correct.
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Re: m03#18 - NS (Polynomial Equations)   [#permalink] 13 Sep 2011, 07:46
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