Question

3x2+4y2-8xy-2x+1= 0 than find (x+ y)2​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$(x+y)^{2} =-2x^{2}-3y^{2}+10xy+2x-1.$

Step-by-step explanation:

We have,

$3x^2+4y^2-8xy-2x+1= 0$

To find, the value of $(x+y)^{2} =?$

$3x^2+4y^2-8xy-2x+1= 0$

⇒ $(x^2+2x^{2} )+(y^2+ 3y^{2} )-8xy-2x+1+2xy-2xy= 0]$

⇒ $(x^2+y^2+2xy)+2x^{2}+3y^{2}-10xy-2x+1= 0$

⇒ $(x+y)^{2} +2x^{2}+3y^{2}-10xy-2x+1= 0$

⇒ $(x+y)^{2} =-2x^{2}-3y^{2}+10xy+2x-1$

Hence, $(x+y)^{2} =-2x^{2}-3y^{2}+10xy+2x-1.$