find the zeroes of the polynomial 7y² - 11/3 y - 2/3
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Answered by
1
Answer:
multiply the eq by 3
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 y = 2/3
a=7 b=-11/3 c=-2/3
sum of roots is -1/7+2/3=11/21 =-b/a
product of root=-2/21=c/a
Answered by
0
Answer:
Step-by-step explanation:
Answer:
2/3 & -1/7
Step-by-step explanation:
Find the zeroes of the polynomial 7y^2 - 11/3 y 2/3 by factorisation method and verify the relations between the zeroes and the coefficients of the polynomial.
7y² - 11y/3 - 2/3 = 0
=> 21y² - 11y - 2 = 0
=> 21y² + 3y - 14y - 2 = 0
=> 3y(7y + 1) - 2(7y + 1) = 0
=> (3y -2)(7y + 1) = 0
=> y = 2/3 or -1/7
Sum of roots = 2/3 - 1/7 = 11/21 = -b/a = - (-11/3)/7 = 11/21
Products of roots = (2/3)(-1/7) = -2/21 = c/a
= (-2/3)/7 =-2/21
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