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if [m][square_root]2x^2 + 2xy + 13y^2[/square_root][/m]
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21 Apr 2020, 11:40
21 Apr 2020, 11:40
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Question Stats:
89% (01:48) correct
10% (02:04) wrong based on 56 sessions
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if \(\sqrt{2x^2 + 2xy + 13y^2}\) \(= x + 3y\) then \(x = \)
A) \(\frac{y}{2}\)
B) \(\frac{y^2}{2}\)
C) \(2y\)
D) \(y-2\)
E) \(y+2\)
A) \(\frac{y}{2}\)
B) \(\frac{y^2}{2}\)
C) \(2y\)
D) \(y-2\)
E) \(y+2\)
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Re: if [m][square_root]2x^2 + 2xy + 13y^2[/square_root][/m]
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01 Sep 2020, 05:39
01 Sep 2020, 05:39
1
workout wrote:
if \(\sqrt{2x^2 + 2xy + 13y^2}\) \(= x + 3y\) then \(x =\)
A) \(\frac{y}{2}\)
B) \(\frac{y^2}{2}\)
C) \(2y\)
D) \(y-2\)
E) \(y+2\)
A) \(\frac{y}{2}\)
B) \(\frac{y^2}{2}\)
C) \(2y\)
D) \(y-2\)
E) \(y+2\)
Take: \(\sqrt{2x^2 + 2xy + 13y^2}= x + 3y\)
Square both sides to get: \((\sqrt{2x^2 + 2xy + 13y^2})^2= (x + 3y)^2\)
Simplify and expand to get: \(2x^2 + 2xy + 13y^2= x^2 + 6xy + 9y^2\)
Subtract \(x^2\) from both sides of the equation to get: \(x^2 + 2xy + 13y^2= 6xy + 9y^2\)
Subtract \(6xy\) from both sides of the equation to get: \(x^2 - 4xy + 13y^2= 9y^2\)
Subtract \(9y^2\) from both sides of the equation to get: \(x^2 - 4xy + 4y^2= 0\)
Factor: \((x-2y)^2 = 0\)
From this we can conclude that: \(x-2y=0\)
Add \(2y\) to both sides to get: \(x=2y\)
Answer: C
Re: if [m][square_root]2x^2 + 2xy + 13y^2[/square_root][/m]
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16 Feb 2024, 14:29
16 Feb 2024, 14:29
Hello from the GRE Prep Club BumpBot!
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
