Question

SOLVE THE QUESTION:-
7y*y-11y/3-2/3​

Answer 1

q(y) = 7y2 – (11/3)y – 2/3  We put q(y) = 0  ⇒ 7y2 – (11/3)y – 2/3 = 0  ⇒  (21y2 – 11y -2)/3 = 0  ⇒ 21y2 – 11y – 2 = 0  ⇒ 21y2 – 14y + 3y – 2 = 0  ⇒ 7y(3y – 2) – 1(3y + 2) = 0  ⇒ (3y – 2)(7y + 1) = 0  This gives us 2 zeros, for  y = 2/3 and y = -1/7  Hence, the zeros of the quadratic equation are 2/3 and -1/7.  Now, for verification  Sum of zeros = – coefficient of y / coefficient of y2  2/3 + (-1/7) = – (-11/3) / 7  -11/21 = -11/21  Product of roots = constant / coefficient of y2  2/3 x (-1/7) = (-2/3) / 7  – 2/21 = -2/21 

q-y-7y-2-11-3-y-2-3

Answer 2

Answer:

7y² - 11/3y -2/3 = 0

Step-by-step explanation:

(21y² -11y -2)/3 =0

21y² -11y -2 =0/3

21y²-11y-2=0

21y²- 14y + 3y -2 =0

7y(3y-2) + 1 (3y-2) =0

(7y+1) (3y-2) =0

Either,

y = -1/7 or 2/3.

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