Chemistry, asked by Dilchor2, 22 days ago


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Answers

Answered by sakshi1158
1

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.

Explanation:

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Answered by mukeshsharma4365
1

Answer:

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