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2x^2+4xy+4y^2=0

 May 15, 2015

Best Answer 

 #3
avatar+23245 
+5

CPhill: I think that you graphed 2x^2 + 4xy + 2y^2 = 0, not 2x^2 + 4xy + 4y^2 = 0

Looking at 2x² + 4xy + 4y²  = 0

Divide by 2:  x² + 2xy + 2y² = 0

Solving for y:        2y² + 2xy =  -x²

Dividing by 2:          y² + xy  = -½x²

Completing the square:  y² + xy + ¼x²  =  -½x² + ¼x²

Factoring:                       (y + ½x)²  =  -¼x²

Solving:                            y + ½x  =  ±√ (-¼x²)

But the radical will be negative unless x = 0

If x = 0,                             y = 0.

When y = 0, x = 0                --->  So, it's only the origin, the point (0, 0).

 May 15, 2015
 #1
avatar+23245 
+5

Do you want to graph this, factor this, or ???

 May 15, 2015
 #2
avatar+128079 
+5

.............................................................

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 May 15, 2015
 #3
avatar+23245 
+5
Best Answer

CPhill: I think that you graphed 2x^2 + 4xy + 2y^2 = 0, not 2x^2 + 4xy + 4y^2 = 0

Looking at 2x² + 4xy + 4y²  = 0

Divide by 2:  x² + 2xy + 2y² = 0

Solving for y:        2y² + 2xy =  -x²

Dividing by 2:          y² + xy  = -½x²

Completing the square:  y² + xy + ¼x²  =  -½x² + ¼x²

Factoring:                       (y + ½x)²  =  -¼x²

Solving:                            y + ½x  =  ±√ (-¼x²)

But the radical will be negative unless x = 0

If x = 0,                             y = 0.

When y = 0, x = 0                --->  So, it's only the origin, the point (0, 0).

geno3141 May 15, 2015
 #4
avatar+128079 
0

Thanks, geno.......my bad.....!!!

 

 

 May 15, 2015

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