Question

Find zeros of following polynomial and verify the relationship between zeros and coefficient.
$({7y}^{2}) - (\frac{11y}{3}) - (\frac{2}{3}) = 0$
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Answer 1

Given polynomial is 7y² - 11y/3 - 2/3 = 0

→ 7y² - 11y/3 - 2/3 = 0

→ (21y² - 11y - 2)/3 = 0

→ 21y² - 11y - 2 = 0

The above equation is in the form ax² + bx + c = 0 or ay² + by + c = 0

Now, we have to find the zeros of the above polynomial by splitting the middle term.

We have to split it in such a way that it's sum is b i.e. (-11) and the product is ac i.e. (-21*2)

→ 21y² - 11y - 2 = 0

→ 21y² - 14y + 3y - 2 = 0

→ 7y(3y - 2) +1(3y - 2) = 0

→ (7y + 1)(3y - 2) = 0

On comparing we get,

→ y = -1/7 and 2/3

Therefore, the zeros are -1/7 and 2/3.

Verification

From above we have polynomial: 21y² - 11y - 2

Here, a = 21, b = -11 and c = -2

Sum of zeros = -b/a

-1/7 + 2/3 = -(-11)/21

(-3 +14)/21 = 11/21

11/21 = 11/21

Product of zeros = c/a

(-1/7) × (2/3) = -2/21

-2/21 = -2/21

Answer 2

||✪✪ QUESTION ✪✪||

• Find zeros of following polynomial and verify the relationship between zeros and coefficient :- 7y² - 11y/3 - 2/3 = 0

|| ✰✰ ANSWER ✰✰ ||

→ 7y² - 11y/3 - 2/3 = 0

Taking LCM in LHS,

→ (7y²*3 - 11y - 2) /3 = 0

→ 21y² - 11y - 2 = 0

Splitting The Middle Term,

→ 21y² + 3y - 14y - 2 = 0

→ 3y(7y + 1) - 2(7y + 1) = 0

→ (7y + 1)(3y - 2) = 0

Putting Both Equal to Zero now,

→ 7y + 1 = 0

→ 7y = (-1)

→ y = (-1)/7

Or,

→ 3y - 2 = 0

→ 3y = 2

→ y = (2/3)

Hence, Zeros of the following polynomial are (-1/7) & (2/3).

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Now, First Relation is :-

→ Sum of Zeros = - (coefficient of y) /(coefficient of y²)

Putting both values :-

→ (-1)/7 + 2/3 = -(-11)/21

→ (-1*3 + 2*7)/21 = (11/21)

→ (-3 + 14)/21 = (11/21)

→ (11/21) = (11/21) ✪✪ Hence Verified. ✪✪

Second Relation :-

→ Product Of Zeros = Constant Term / (coefficient of y²)

Putting both Values :-

→ (-1/7) * (2/3) = (-2)/21

→ (-2)/21 = (-2)/21 ✪✪ Hence Verified. ✪✪

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