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505-555 (Easy)|   Algebra|      
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Bunuel
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 19 Mar 2015, 06:24
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C
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15% (low)

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84% (02:05) correct 16% (02:50) wrong based on 275 sessions

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If \(\sqrt{2x^2 + 2xy + 13y^2}= x + 3y\) , then x =

A. y/2
B. y^2/2
C. 2y
D. y - 2
E. y + 2

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C
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 20 Mar 2015, 09:57
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Bunuel
If \(\sqrt{2x^2 + 2xy + 13y}= x + 3y\) , then x =

A. y/2
B. y^2/2
C. 2y
D. y - 2
E. y + 2

Kudos for a correct solution.
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hi bunuel,

i have not solved the equation but looking at the equation, coefficients and the choices, the answer does not seem to be there..
reasons...
\(\sqrt{2x^2 + 2xy + 13y}= x + 3y\)......or 2x^2 + 2xy + 13y= \((x + 3y)^2\)
1) Since any value of x only in term of y or multiple of y will not give us a y to the power 1 on RHS to equal 13y on the LHS... so A,B and C are out..
2) if we look at D and E, the remaining values of x is in terms of y with a number 2 added or subtracted. this makes RHS coefficient of \(y^2\) as 16, which does not equal 4 on LHS..

i may be wrong as i say this looking at the coefficients and power of variables... i'll ofcourse try to solve it by mathematical way now..
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 20 Mar 2015, 10:13
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chetan2u
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If \(\sqrt{2x^2 + 2xy + 13y}= x + 3y\) , then x =

A. y/2
B. y^2/2
C. 2y
D. y - 2
E. y + 2

Kudos for a correct solution.


hi bunuel,

i have not solved the equation but looking at the equation, coefficients and the choices, the answer does not seem to be there..
reasons...
\(\sqrt{2x^2 + 2xy + 13y}= x + 3y\)......or 2x^2 + 2xy + 13y= \((x + 3y)^2\)
1) Since any value of x only in term of y or multiple of y will not give us a y to the power 1 on RHS to equal 13y on the LHS... so A,B and C are out..
2) if we look at D and E, the remaining values of x is in terms of y with a number 2 added or subtracted. this makes RHS coefficient of \(y^2\) as 16, which does not equal 4 on LHS..

i may be wrong as i say this looking at the coefficients and power of variables... i'll ofcourse try to solve it by mathematical way now..
Show more

There was a typo (13y instead of 13y^2). Edited. Thank you.
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 20 Mar 2015, 10:29
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Bunuel
If \(\sqrt{2x^2 + 2xy + 13y^2}= x + 3y\) , then x =

A. y/2
B. y^2/2
C. 2y
D. y - 2
E. y + 2

Kudos for a correct solution.
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hi all,

although this question may take some time in finding an answer, a quick method to these slightly complex Q is matching the coefficients...
lets see the question...
\(\sqrt{2x^2 + 2xy + 13y^2}= x + 3y\)....or \(2x^2 + 2xy + 13y^2=( x + 3y)^2\)

lets do elimination by looking at coefficients...
1) if we take x in terms of y/2 or y^2/2, the RHS will have some coefficient of y^2 or y^4 with 4 in denominator but the LHS will have a 2 in denominator...
so we can eliminate A and B..
2) if we take values of x is in terms of y with a number 2 added or subtracted, this makes RHS coefficient of y^2 as 16, which does not equal 4 on LHS.
so we can eliminate D and E..
3) we are now left with only C, which should be the answer.. lets check..
\(2x^2 + 2xy + 13y^2=( x + 3y)^2\)...
\(2(2y)^2 + 2*2y*y + 13y^2=( 2y + 3y)^2\)....
\(8y^2 + 4y^2 + 13y^2=( 2y + 3y)^2\)..
\(25 y^2=25y^2\)...
so ans is C..2y
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 20 Mar 2015, 11:17
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Answer C.

Upon simplifying, we get, x2+4y2=4xy

-> X2-4xy+4y2=0

(x-2y)2=0

X-2y=0

X=2y.

-

Bunuel
If \(\sqrt{2x^2 + 2xy + 13y^2}= x + 3y\) , then x =

A. y/2
B. y^2/2
C. 2y
D. y - 2
E. y + 2

Kudos for a correct solution.
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 20 Mar 2015, 11:52
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answer is C

solve the equation and equate x to y and find the relation

-h
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 20 Mar 2015, 20:19
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Bunuel
If \(\sqrt{2x^2 + 2xy + 13y^2}= x + 3y\) , then x =

A. y/2
B. y^2/2
C. 2y
D. y - 2
E. y + 2

Kudos for a correct solution.
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I was starting to think that this was a really difficult problem before the edit.

if we square both sides,

(x^2)-4xy+4(y^2) = 0
(x-2y)(x-2y)=0

x-2y=0
x=2y

Answer: C
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 20 Mar 2015, 20:31
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Yes C is the right option.

Posted from my mobile device
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 21 Mar 2015, 07:08
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I looked at it, wanted to run , then decided to plug in answer choices

I plugged in C and got it right, so basically, whenever the question looks very scary, calm down and test answer choices where possible
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 23 Mar 2015, 00:03
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Answer = C = 2y

Squaring both sides

\(2x^2 + 2xy + 13y^2 = (x+3y)^2\)

\((x-2y)^2 = 0\)

x = 2y
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 23 Mar 2015, 05:50
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Bunuel
If \(\sqrt{2x^2 + 2xy + 13y^2}= x + 3y\) , then x =

A. y/2
B. y^2/2
C. 2y
D. y - 2
E. y + 2

Kudos for a correct solution.
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MAGOOSH OFFICIAL SOLUTION:
Attachment:
variablesroots_explanation.png
variablesroots_explanation.png [ 30.37 KiB | Viewed 13062 times ]
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 01 Sep 2020, 06:38
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Bunuel
If \(\sqrt{2x^2 + 2xy + 13y^2}= x + 3y\) , then x =

A. y/2
B. y^2/2
C. 2y
D. y - 2
E. y + 2

Kudos for a correct solution.
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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
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If (2x^2 + 2xy + 13y^2)^(1/2) = x + 3y , then x = 03 Jul 2024, 11:41
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