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by [email protected] » Wed Jan 08, 2014 5:10 am
I did this question by taking x^2=X therefore

the equation will be x^2-2x+1 to find the roots - X^2-1x-1x+1=0 therefore the roots will be

x(x-1)-1(x-1) we get x^2=1 and x^2=1 aren't we getting one root as the answer ?

What am I doing wrong here?
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by Brent@GMATPrepNow » Wed Jan 08, 2014 6:32 am
Your work is perfect right up until you determined that x^2=1
At this point, there are 2 possible solutions: x = 1 and x = -1

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by Brent@GMATPrepNow » Wed Jan 08, 2014 7:07 am
How many distinct roots does the equation x� - 2x² + 1 = 0 have?
A) 0
B) 1
C) 2
D) 3
E) 4
Here's my solution.
First notice that x� = (x²)²

Now, we'll use something called "u-substitution"

Let u = x²
So, we can take the equation replace x² with u to get: u² - 2u + 1 = 0
Factor to get: (u - 1)(u - 1) = 0
Solve, to get u = 1

Now that we know that u = 1, we'll recognize that u = x²
So, we now know that x² = 1

If x² = 1, then we have two possible solutions: x = 1 and x = -1
Since the equation has two distinct roots (solutions), the correct answer is C

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Brent
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by gmaster328 » Thu Jan 09, 2014 8:27 pm
I think, because of the even exponent, there are two roots, not one
[email protected] wrote:I did this question by taking x^2=X therefore

the equation will be x^2-2x+1 to find the roots - X^2-1x-1x+1=0 therefore the roots will be

x(x-1)-1(x-1) we get x^2=1 and x^2=1 aren't we getting one root as the answer ?

What am I doing wrong here?
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by Matt@VeritasPrep » Sun Jan 12, 2014 6:29 pm
As Brent says,

x� - 2x² + 1 = 0

can be factored as

(x² - 1)(x² - 1) = 0

which gives us

(x² - 1) = 0

or

x² = 1

So we have two roots, x = 1 and x = -1.

The tricky thing here is that x² = 1 has two solutions (whereas, on the GMAT, √1 only has one solution). Mathematical conventions can be a pain!
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