find the zeros of
7y^2-11/3-2/3
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2
Answer:
2 is the correct answer
Answered by
4
Answer:
Given,
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
Therefore, the relationship between zeros and their coefficients is verified..
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