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Its B

2x^2 -y^2 = 2xy
2x^2 = 2xy + y^2

Add x^2 on both sides we get

2x^2 +x^2 = x^2 +y^2 +2xy
3x^2 = (x+y)^2
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bakfed
If 2x^2 - y^2 = 2xy, then (x+y)^2 =

A. x^2
B. 3x^2
C. 4xy
D. 2y^2
E. -y^2

OA
here we have to find the value of x^2 +y^2 +2xy in which 2xy
=2x^2 -y^2 (given ) hence ans is 3x^2.
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2x^2 –y^2 = 2xy
2x^2 = 2xy+y^2

3x^2 = x^2 + 2xy+y^2 (adding x^2 on both sides)

3x^2 = (x+y)^2


Answer:- B
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2x\(^2\)-y\(^2\)=2xy

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

replace 2xy by 2x\(^2\)-y\(^2\) and solve

Answer B.
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Sarang
2x^2 –y^2 = 2xy
2x^2 = 2xy+y^2

3x^2 = x^2 + 2xy+y^2 (adding x^2 on both sides)

3x^2 = (x+y)^2


Answer:- B

i didn't understand this pls explain
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Sarang
2x^2 –y^2 = 2xy
2x^2 = 2xy+y^2

3x^2 = x^2 + 2xy+y^2 (adding x^2 on both sides)

3x^2 = (x+y)^2


Answer:- B

i didn't understand this pls explain

Given: \(2x^2-y^2=2xy\) --> rearrange: \(2x^2=2xy+y^2\) --> add \(x^2\) to both parts of the equation: \(2x^2+x^2=x^2+2xy+y^2\) --> \(3x^2=(x+y)^2\).

Hope it's clear.
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\(2x^2 – y^2 = 2xy\)

so, \(y^2 = 2x^2 - 2xy\) ............ (1)

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

Placing value of \(y^2\) from (1)

\(= x^2 + 2xy + 2x^2 - 2xy\)

\(= 3x^2\)

Answer = B
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Quite straight forward question

step 1:
(x+Y)^2 = X^2 +2xy + y^2

step 2:
now substitute 2xy from the equation.

(x+Y)^2 = X^2 +2x^2-y^2+y^2.

-y^2 and Y^2 cancels out

we are left with x^2+2X^2= 3X^2 Hence B.
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bakfed
If 2x^2 - y^2 = 2xy, then (x+y)^2 =

A. x^2
B. 3x^2
C. 4xy
D. 2y^2
E. -y^2

We see that we are looking for the value of (x + y)^2 = x^2 + y^2 + 2xy.

We also see that 2x^2 - y^2 = 2xy; thus we can substitute for 2xy and we have:

x^2 + y^2 + 2x^2 - y^2 = 3x^2

Answer: B
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Given: 2xy=2x2−y22xy=2x2−y2.

(x+y)2=x2+2xy+y2(x+y)2=x2+2xy+y2
substitute 2xy2xy
(x+y)2=x2+(2x2−y2)+y2=3x2(x+y)2=x2+(2x2−y2)+y2=3x2.

B.
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\(2x^2 - y^2 = 2xy\) ----------Equation (1)

=> \((x + y)^2 = x^2 + y^2 + 2xy\) ----------Equation (2)

Multiply Equation (1) by (-1),

=> \(- 2x^2 + y^2 = -2xy \)

=> \(- 2x^2 + y^2 + 2xy = 0\)

=> \(- 3x^2 + x^2 + y^2 + 2xy = 0 \)

From ----------Equation (2),

=> \(- 3x^2 + (x + y)^2 = 0 \)

=> \((x + y)^2 = 3x^2 \)

Answer B
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Given: \(2x^2 - y^2 = 2xy\)

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

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

\(= 3x^2\\
\)

Answer is B.
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