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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If [m]x^2-4xy+4y^2=0[/m] and [m]x^n = y^n[/m], then which of
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20 Aug 2019, 09:59
20 Aug 2019, 09:59
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\(xy≠0\), \(x^2-4xy+4y^2=0\) and \(x^n = y^n\).
Which of the following COULD be the value of \(n\)?
i) 0
ii) 1
iii) 2
A) i only
B) ii only
C) i & ii only
D) ii & iii only
E) i, ii, & iii
Which of the following COULD be the value of \(n\)?
i) 0
ii) 1
iii) 2
A) i only
B) ii only
C) i & ii only
D) ii & iii only
E) i, ii, & iii
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If [m]x^2-4xy+4y^2=0[/m] and [m]x^n = y^n[/m], then which of
[#permalink]
20 Aug 2019, 10:07
20 Aug 2019, 10:07
1
GreenlightTestPrep wrote:
\(xy≠0\), \(x^2-4xy+4y^2=0\) and \(x^n = y^n\). Which of the following COULD be the value of \(n\)?
i) 0
ii) 1
iii) 2
A) i only
B) ii only
C) i & ii only
D) ii & iii only
E) i, ii, & iii
i) 0
ii) 1
iii) 2
A) i only
B) ii only
C) i & ii only
D) ii & iii only
E) i, ii, & iii
USEFUL PROPERTY : \(\frac{x^k}{y^k}=(\frac{x}{y})^k\)
Given: \(n\) is an integer, \(xy ≠ 0\), and \(x^n = y^n\)
Take: \(x^n = y^n\)
Since \(y ≠ 0\), we can safely divide both sides by \(y^n\) to get: \(\frac{x^n}{y^n}=1\)
Apply above PROPERTY to get: \((\frac{x}{y})^n=1\)
Now take the other given information: \(x^2-4xy+4y^2=0\)
Factor to get: \((x-2y)(x-2y)=0\)
So, we can conclude that \(x-2y=0\)
Add \(2y\) to both sides to get: \(x=2y\)
Divide both sides by \(y\) to get: \(\frac{x}{y}=2\)
From here, can take \((\frac{x}{y})^n=1\) and replace \(\frac{x}{y}\) with \(2\) to get: \(2^n=1\)
This means that \(n=0\)
Since no other values of \(n\) satisfy the equation \(2^n=1\), the correct answer is A
Cheers,
Brent
Re: If [m]x^2-4xy+4y^2=0[/m] and [m]x^n = y^n[/m], then which of
[#permalink]
22 Aug 2019, 14:07
22 Aug 2019, 14:07
1
GreenlightTestPrep wrote:
\(xy≠0\), \(x^2-4xy+4y^2=0\) and \(x^n = y^n\).
Which of the following COULD be the value of \(n\)?
i) 0
ii) 1
iii) 2
A) i only
B) ii only
C) i & ii only
D) ii & iii only
E) i, ii, & iii
Which of the following COULD be the value of \(n\)?
i) 0
ii) 1
iii) 2
A) i only
B) ii only
C) i & ii only
D) ii & iii only
E) i, ii, & iii
Here's how I did this one.
Factor the polynomial which yieds (x-2y)(x-2y) = (x-2y)^2 = 0 which isn't to hard.
so x = 2y.
If x and 2y are by definition different values (since we can't let either one equal 0) then it becomes
clear that if they're getting raised to the same power the only way to get them equal is to raise them to the 0.
Thus the answer is A.
