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If $(4x + 2y)^2 = 64$ and $xy = 2$, what is the value of $4x^2 + y^2$?

 Jun 14, 2019
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(4x + 2y)2  =  64

                                             Expand the left side of the equation.

(4x + 2y)(4x + 2y)  =  64

 

16x2 + 16xy + 4y2  =  64

                                             Divide both sides of the equation by  4

4x2 + 4xy + y2  =  16

                                             Subtract  4xy  from both sides of the equation.

4x2 + y2  =  16 - 4xy

                                             Since  xy = 2  we can substitute  2  in for  xy

4x2 + y2  =  16 - 4(2)

                                             Simplify the right side of the equation.

4x2 + y2  =  8

 Jun 15, 2019

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