Find the zeroes of the quadratic polynomial 7y^2 - 11/3 y - 2/3 and verify the relationship between the zeroes and the coefficients

Question

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.

Answer 1

Given, the polynomial is 7y² - (\(\frac{1}{3}\))y - (2/3).

The polynomial can be rewritten as (\(\frac{1}{3}\))[21y² - 11y - 2]

Let (\(\frac{1}{3}\))[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 = \(-\frac{1}{7}\)

Also, 3y - 2 = 0

3y = 2

y = \(\frac{2}{3}\)

Therefore, the zeros of the polynomial are \(\frac{2}{3}\) and \(-\frac{1}{7}\).

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

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

Product of the roots is αβ = 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 + β

= -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: αβ

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

= -2/21

RHS: constant term/coefficient of x²

= -2/21

LHS = RHS