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7y² - (11/3)y - (2/3). Find the zeroes of the polynomial , and verify the relation between the coefficients and the zeroes of the polynomial

Solution:

Given, the polynomial is 7y² - (11/3)y - (2/3).

We have to find the relation between the coefficients and zeros of the polynomial

The polynomial can be rewritten as (1/3)[21y² - 11y - 2]

Let (1/3)[21y² - 11y - 2] = 0

21y² - 11y - 2 = 0

On factoring,

21y² - 14y + 3y - 2 = 0

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

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

Now, 7y + 1 = 0

7y = -1

y = -1/7

Also, 3y - 2 = 0

3y = 2

y = 2/3

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

We know that, if 𝛼 and ꞵ are the zeroes of a polynomial ax² + bx + c, then

Sum of the roots is 𝛼 + ꞵ = -coefficient of x/coefficient of x² = -b/a

Product of the roots is 𝛼ꞵ = constant term/coefficient of x² = c/a

From the given polynomial,

coefficient of x = -11

Coefficient of x² = 21

Constant term = -2

Sum of the roots:

LHS: 𝛼 + ꞵ

= -1/7 + 2/3

= (-3+14)/21

= 11/21

RHS: -coefficient of x/coefficient of x²

= -(-11)/21

= 11/21

LHS = RHS

Product of the roots

LHS: 𝛼ꞵ

= (-1/7)(2/3)

= -2/21

RHS: constant term/coefficient of x²

= -2/21

LHS = RHS

✦ Try This: Find the zeroes of the polynomial x² + (3/5)x - 8, and verify the relation between the coefficients and the zeroes of the polynomial

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 2

NCERT Exemplar Class 10 Maths Exercise 2.3 Problem 10

7y² - (11/3)y - (2/3). Find the zeroes of the polynomial , and verify the relation between the coefficients and the zeroes of the polynomial

Summary:

The zeroes of the polynomial 7y² - (11/3)y - (2/3) are -1/7 and 2/3. The relation between the coefficients and zeros of the polynomial are, Sum of the roots = -b/a = 11/21, Product of the roots = c/a = -2/21

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