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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]

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New post 22 Apr 2006, 18:07
How many solutions does equation (x^2-25)^2=x^2-10x+25 have?

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Re: squared equations [#permalink]

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New post 22 Apr 2006, 18:29
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]

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New post 23 Apr 2006, 01:31
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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New post 23 Apr 2006, 08:19
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]

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New post 23 Apr 2006, 08:25
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]

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New post 23 Apr 2006, 08:30
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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New post 24 Apr 2006, 05:59
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]

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New post 07 Feb 2014, 21:13
(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]

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New post 10 Feb 2014, 09:25
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.

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Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink]

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New post 11 Feb 2014, 12:34
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.

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Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink]

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New post 22 May 2014, 08:12
LokeshB wrote:
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.


Dear Lokesh

You are making a fundamental mistake.

If it's given that \(a^2 = b^2\)
This implies that,\(a = +b\) or \(a= -b\)

So, in the question above, when you took the square root on both sides, you would get two equations:

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

AND

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

Hope this helped! :)
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Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink]

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New post 23 May 2014, 07:43
LokeshB wrote:
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.


Hi Lokesh,

You are missing out on the solution of LHS = RHS = 0
Taking x -5 as a root, we get x =5
And putting x = 5 in the equation we get LHS = RHS = 0

Hence, x = 5 is one of the solutions

Golden Rule: Be careful while cancelling out factors involving variables

Rgds,
Rajat
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Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink]

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New post 23 May 2014, 11:10
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This is a perfect example of an official GMAT question because it looks complicated but is meant to be solved solely by factoring and using the difference of squares identity. Of course one could expand the entire expression and create a mess, but GMAT writers do not expect students to go in that direction.

See the attached solution in the image.

Cheers,
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Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink]

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New post 12 Jun 2014, 22:55
Solution: 2
x = -4, -6
(x^2-25)^2=x^2-10x+25
{(x+5)(x-5)}^2=(x-5)^2
(x+5)^2 X (x-5)^2=(x-5)^2
(x+5)^2=1
x^2+10x+25=1
x^2+10x+24=0
x=-4,-6
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Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink]

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New post 13 Jun 2014, 08:05
@aknine

{(x+5)(x-5)}^2=(x-5)^2

At this stage, we cannot divide both sides by (x-5)^2, because that would mean ignoring the solution x=5.
Instead, subtract and factor, and that would show that x=5 is also a solution.

Also, once you are at the next stage of
(x+5)^2=1
there is no need expand the (x+5)^2 term, instead we can directly conclude:

x+5 = 1 or x+5=-1
which gives the remaining two solutions of x=4 and x = -6.

Cheers,
Dabral

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Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have   [#permalink] 13 Jun 2014, 08:05
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