# If alpha and beta are the zeroes of the polynomial 6y 2 -7y +2 Find the quadratic polynomial whose zeroes are 1/alpha and 1/beta

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a+b=-(-7)/6=7/6. .....(1)

ab=2/6=1/3. .....(2)

Now,the required equation has roots: 1/a & 1/b.

sum of roots,S=(1/a)+(1/b)

=(a+b)/ab

=(7/6)÷(1/3). {from 1 & 2}

S =7/2. ...(3)

product of roots,P=(1/a)×(1/b)

=1/(ab)

=1÷(1/3)

P =3. ...(4)

The required equation will be of the form:

y^2 -Sy+P = y^2-(7/2)y+3=O [from 3 & 4]

=2y^2-7y+6=0

Hence,the required equation is:

2y^2 - 7y + 6 = 0

6y²-4y-3y+2

2y(3y-2) - 1(3y-2)

(3y-2)(2y-1)

Zeroes = 2/3 or 1/2

If alpha = 2/3 and beta = 1/2

then 1/alpha = 3/2 and 1/beta = 2

Now the quadratic polynomial is

x²-(sum of roots)x + (product of the roots)

Sum of the roots = 3/2 + 2 = 3.5

product of the roots = 3/2 * 2 = 3

The quadratic is

x² - 3.5x + 3