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

Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink]
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 \(x = 6\)

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

1

This post received KUDOS

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, Dabral

Attachments

GMATClub-05232014.png [ 62 KiB | Viewed 699 times ]

Re: How many solutions does equation (x^2-25)^2=x^2-10x+25 have [#permalink]
13 Jun 2014, 07:05

@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.

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