Question

Find the zeroes of the quadratic polynomial 7y2 -11/3 y-2/3 and verify the relationship between the zeroes and the coefficients. i need explanation ​

Answer 1

Answer:

Let the zeros of the given equation be α and β.

Now,

7y² - 11/3 y - 2/3 = 0

On multiplying both sides by 3, we get,

21y² - 11y - 2 = 0

21y² - 14y + 3y - 2 = 0

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

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

⇒ Either (7y + 1) = 0 or (3y - 2) = 0

∴ y = -1/7 or 2/3

∴ α = -1/7 & β = 2/3

Now, the equation is 7y² - 11/3 y - 2/3 = 0.

Here, a = 7, b = -11/3, c = -2/3, α = -1/7 & β = 2/3.

Now,

Sum of the roots = α + β = -b/a

LHS = α + β = 11/21

RHS = -b/a = 11/21

∴ LHS = RHS

Now,

Product of the roots = αβ = c/a

LHS = αβ = -2/21

RHS = c/a = -2/21

∴ LHS = RHS

Hence, verified.

Hope it helps!!!

Answer 2

Answer:

Follow these stepsownward Let the zeros of the given equation be α and β.

Now,

7y² - 11/3 y - 2/3 = 0

On multiplying both sides by 3, we get,

21y² - 11y - 2 = 0

21y² - 14y + 3y - 2 = 0

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

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

⇒ Either (7y + 1) = 0 or (3y - 2) = 0

∴ y = -1/7 or 2/3

∴ α = -1/7 & β = 2/3

Now, the equation is 7y² - 11/3 y - 2/3 = 0.

Here, a = 7, b = -11/3, c = -2/3, α = -1/7 & β = 2/3.

Now,

Sum of the roots = α + β = -b/a

LHS = α + β = 11/21

RHS = -b/a = 11/21

∴ LHS = RHS

Now,

Product of the roots = αβ = c/a

LHS = αβ = -2/21

RHS = c/a = -2/21

∴ LHS = RHS

Hence, verified.

Hope it helps!!!

Step-by-step explanation: