Question

find the roots of the quadrantic equation 9y2-3y=2​

Answer 1

Answer:

it can be written as :-

9y²-3y -2 = 0

9y² -6y + 3y -2 = 0

3y(3y-2) +1(3y-2) = 0

( 3y + 1 ) ( 3y - 2 ) = 0

$<marquee><font color="red">the zeros are -1/3 and +2/3 </marquee>$

Answer 2

9y2-3y = 2

9y2-3y-2=0

9y2-3y+6y-2=0

3y(3y-1) + 2(3y-1)=0

(3y+3), (3y-2)

So, y = 1, 2/3