**Quadratic Polynomial with Zeroes 1/alpha and 1/beta**

To find a quadratic polynomial whose zeroes are 1/alpha and 1/beta, we need to use the relationship between the zeroes and coefficients of a polynomial.

**Relationship Between Zeroes and Coefficients**

For a quadratic polynomial in the form of ax^2 + bx + c, the sum of the zeroes is given by:

alpha + beta = -b/a (1)

The product of the zeroes is given by:

alpha * beta = c/a (2)

**Given Polynomial**

The given polynomial is:
6y^2 - 7y

Let's find the sum and product of the zeroes using the given polynomial.

**Finding the Sum and Product of Zeroes**

Comparing the given polynomial with the general form, we can equate the coefficients as follows:

a = 6
b = -7
c = 0 (since there is no constant term)

Using equations (1) and (2), we can now find the sum and product of the zeroes:

alpha + beta = -(-7)/6 = 7/6
alpha * beta = 0/6 = 0

Thus, the sum of the zeroes is 7/6 and the product of the zeroes is 0.

**New Quadratic Polynomial**

To find a new quadratic polynomial with zeroes 1/alpha and 1/beta, we need to use the relationship between the zeroes and coefficients.

Let's consider the new quadratic polynomial as ay^2 + by + c.

Using equations (1) and (2), we can substitute the values of the sum and product of the zeroes:

1/alpha + 1/beta = -b/a
1/(alpha * beta) = c/a

Substituting the values, we have:

1/(1/alpha) + 1/(1/beta) = -b/a
alpha + beta = -b/a

Since we know that alpha + beta = 7/6, we can equate it to -b/a:

7/6 = -b/a

Simplifying the equation, we have:

b = -7a/6

Similarly, substituting the value of the product of the zeroes:

1/(alpha * beta) = c/a
1/0 = c/6

Since the product of the zeroes is 0, we can equate it to c/6:

0 = c/6

Simplifying the equation, we have:

c = 0

Therefore, the new quadratic polynomial with zeroes 1/alpha and 1/beta is:

ay^2 + (-7a/6)y + 0

Simplifying further, we have:

ay^2 - 7ay/6

Thus, the quadratic polynomial with zeroes 1/alpha and 1/beta is ay^2 - 7ay/6.

From given polynomial we have two zeroes 2/3 and 1/2 as alpha and Bita respective (Suppose)Now For Finding polynomialá=alpha=1/(2/3)=3/2B=Bita=1/(1/2)=2so,Polynomial=y^2-(á+B)y+áB=0y^2-{(3/2)+2}+{(3/2)×2}=0(2y^2-7y+6)/2=0i.e.2y^2-7y+6=0