y²+1/y =2 how to solve and say that its a quadratic equation
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Answered by
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y²+1/y= 2
y³+1/y= 2
y³+1 = 2y
y³-2y+1= 0
y³+y+y+1= 0
y(y²+1)+1(y+1)=0
(y+1)(y²+1)= 0
y= -1 , √1
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Answered by
1
Answer:
(y^2+1)/y = 2
so (y^2 + 1) = 2y
so y^2 - 2y + 1 = 0
so y^2 - y - y + 1 = 0
y(y - 1) - 1(y - 1) = 0
(y-1)(y-1) = 0
Therefore y - 1 = 0
so y = 1
Since degree of this equation is 2, thus we can say it is quadratic equation
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