what is the answer of this Q 2y²+3y+1
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1
2y² + 3y + 1 = 0
2y² + 2y + 1y + 1 = 0
2y ( y + 1) + 1 ( y + 1) = 0
(2y+1) (y+ 1) = 0
either 2y + 1 = 0 or y+ 1 = 0
i. e, y = – 1/2 or y = –1
Answered by
2
product=2
sum=3
the number we will take =2 and 1
2y^2+2y+y+1
now we will take common
2y(y+1) 1(y+1)
=(2y+1) (y +1)
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