Question

Find the zeroes of 7y^2-11/3y-2/3 and also find the relationship between the zeroes of the polynomial and their coefficient...
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Answer 1

The general form of quadratic equation is
ax² + b x+ c = 0


7y²-11/3y-2/3 = 0
21y²-11y-2 = 0
21y²-14y+3y-2=0
7y(3y-2)+1(3y-2) = 0
(7y+1)(3y-2)=0
y=-1/7 and 2/3

sum of zeroes = -1/7 + 2/3
= -3+14/21
= 11/21
= -b/a

product of zeros = (-1/7)(2/3)
= -2/21
= c/a

here is ur answer

i hope it helps u

Answer 2

Let p(y) = 7y2 − 11y3 − 2/3
Now, zeroes of p(y) are gi en by,      p(y) = 0
⇒7y2 − 11y3 − 2/3 = 0
⇒21y2 − 11y − 2 = 0
⇒21y2−14y+3y−2 = 0
⇒7y(3y−2)+1(3y−2) =0
⇒(7y+1)(3y−2) = 0
⇒7y+1 = 0   or   3y−2 = 0
⇒y = −17  or  y = 23
Sum of zeroes = −17+23 = −3+14/21 = 11/21 
= −coefficient of ycoefficient of y2Product of zeroes =−1/7×2/3 =−2/21 = constant termcoefficient of y2