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How many solutions does equation (x^2-25)^2=x^2-10x+25 have

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How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink] New post 22 Apr 2006, 17:07
How many solutions does equation (x^2-25)^2=x^2-10x+25 have?
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Re: squared equations [#permalink] New post 22 Apr 2006, 17:29
joemama142000 wrote:
How many solutions does equation (x^2-25)^2=x^2-10x+25 have?


Using the fundamental theorem of algebra I would say 4, but not all of them necesarily real or different.
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Re: squared equations [#permalink] New post 23 Apr 2006, 00:31
conocieur wrote:
Using the fundamental theorem of algebra I would say 4, but not all of them necesarily real or different.


That's correct. In this case, all are real, but we have twice +5.
(x^2 - 25)^2 = x^2 - 10x + 25
x^4 - 51x^2 - 10x + 600 = 0
(x+5)(x+5)(x-4)(x-6) = 0

x = - 5 OR x = +4 OR x = +6
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 [#permalink] New post 23 Apr 2006, 07:19
the answer is right , but the solution is wrong
try to substitute your answers into the actual equation...


my solution is:
(x - 5 ) ( x - 5 ) ( x + 6 ) ( x +4 ) = 0
x = 5
x = -6
x = -4
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Re: squared equations [#permalink] New post 23 Apr 2006, 07:25
A different approach :

(x^2 - 25)^2 = (x-5)^2

We get 2 equations x^2 - 25 = x-5 and x^2 - 25 = -(x-5)

From 1 we get x^2 -x -20 = (x-5)(x+4)

From 2 we get x^2+x-30 =(x+6)(x-5)

Hence 3 solutions 5,-4 and -6

But my answers are different from yours..ccax
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Re: squared equations [#permalink] New post 23 Apr 2006, 07:30
sherinaparvin wrote:
But my answers are different from yours..ccax

You're right. I swopped the sign for each of the solutions.
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 [#permalink] New post 24 Apr 2006, 04:59
Ccax please elaborate

from
x^4 - 51x^2 - 10x + 600 = 0

how you received this one
(x+5)(x+5)(x-4)(x-6) = 0?

And there should be +10x, not -10x.
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Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink] New post 07 Feb 2014, 20:13
(x^2-25)^2=x^2-10x+25
=> |x^2-25|=|x-5|
Case 1: Both +ve or both negative: x^2-25 = x-5 => x^2 - x -20 = 0 => (x-5)(x+4) = 0 => x = 5, -4
Case 2: One +ve and other -ve: x^2-25 = -(x-5) = -x + 5 => x^2 + x -30 = 0 => (x+6)(x-5)= 0 => x = -6, 5

Roots: -6, -4, 5.
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Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink] New post 10 Feb 2014, 08:25
The best solution here is to put everything on the same side:

(x^2 - 25)^2 = x^2 - 10x + 25
x^4 - 51x^2 - 10x + 600 = 0 => (x+5)(x+5)(x-4)(x-6) = 0
x = - 5 or it could be x = 4 or it could be x = 6

The answer is 3.

hih
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Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink] New post 11 Feb 2014, 11:34
Where am i going wrong? Can someone explain:

Given: (x^2 - 25)^2=x^2-10x+25
=>(x^2 - 25)^2 = (x-5)^2
=>(x^2 - 25) = (x-5)
=>(x^2 - 5^2) = (x-5)
=>(x-5)(x+5) = (x-5)
=>(x+5) = 1
=> x = -4
Just one solution.

Thanks.
Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have   [#permalink] 11 Feb 2014, 11:34
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