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If k and -k are the solutions to x^4 - 10x^2 + 25 = 0, where k is a po

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If k and -k are the solutions to x^4 - 10x^2 + 25 = 0, where k is a po [#permalink]

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New post 01 Feb 2017, 09:36
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Re: If k and -k are the solutions to x^4 - 10x^2 + 25 = 0, where k is a po [#permalink]

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New post 01 Feb 2017, 20:49
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Solutions of x^4 - 10x^2 + 25 = 0 are √k and - √k;

So we can say x = √k and x = - √k and x^2 = k;

Now,
x^4 - 10x^2 + 25 = 0
=> (x^2 -5)^2 = 0
=> x^2 -5 = 0
=> x^2 = 5
=> k = 5; Hence, ans D.


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Re: If k and -k are the solutions to x^4 - 10x^2 + 25 = 0, where k is a po [#permalink]

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New post 02 Feb 2017, 08:14
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Bunuel wrote:
If \(\sqrt{k}\) and \(-\sqrt{k}\) are the solutions to x^4 - 10x^2 + 25 = 0, where k is a positive constant, then k =

A. 0
B. 1
C. 3
D. 5
E. 10


Let's first solve the equation x⁴ - 10x² + 25 = 0
Factor to get: (x² - 5)(x² - 5) = 0
So, it must be the case that x² - 5 = 0
Take x² - 5 = 0 and add 5 to both sides to get: x² = 5

So, x = √5 or x = -√5
We're told that the solutions are √k and -√k
This means that k = 5

Answer: D

Cheers,
Brent
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Re: If k and -k are the solutions to x^4 - 10x^2 + 25 = 0, where k is a po [#permalink]

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New post 08 Feb 2017, 10:35
Bunuel wrote:
If \(\sqrt{k}\) and \(-\sqrt{k}\) are the solutions to x^4 - 10x^2 + 25 = 0, where k is a positive constant, then k =

A. 0
B. 1
C. 3
D. 5
E. 10


Let’s start by factoring x^4 - 10x^2 + 25 = 0:

(x^2 - 5)(x^2 - 5) = 0

Thus:

x^2 - 5 = 0

x^2 = 5

x = √5 or x = -√5, and thus k = 5.

Answer: D
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Re: If k and -k are the solutions to x^4 - 10x^2 + 25 = 0, where k is a po   [#permalink] 08 Feb 2017, 10:35
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