#### Question

If α and β are the zeros of the quadratic polynomial f(x) = x^{2} − 3x − 2, find a quadratic polynomial whose zeroes are `1/(2alpha+beta)+1/(2beta+alpha)`

#### Solution

Since α and β are the zeros of the quadratic polynomial f(x) = x^{2} − 3x − 2

The roots are α and β

`alpha+beta="-coefficient of x"/("coefficient of "x^2)`

`alpha+beta=-((-3)/1)`

α + β = -(-3)

α + β = 3

`alphabeta="constant term"/("coefficient of "x^2)`

`alphabeta=(-2)/1`

αβ = -2

Let S and P denote respectively the sum and the product of zero of the required polynomial . Then,

`S=1/(2alpha+beta)+1/(2beta+alpha)`

Taking least common factor then we have ,

`S=1/(2alpha+beta)xx(2beta+alpha)/(2beta+alpha)+1/(2beta+alpha)xx(2alpha+beta)/(2alpha+beta)`

`S=(2beta+alpha)/((2alpha+beta)(2beta+alpha))+(2alpha+beta)/((2beta+alpha)(2alpha+beta))`

`S=(2beta+alpha+2alpha+beta)/((2alpha+beta)(2beta+alpha))`

`S=(3beta+3alpha)/(4alphabeta+2beta^2+2alpha^2+betaalpha)`

`S=(3(beta+alpha))/(5alphabeta+2(alpha^2+beta^2))`

`S=(3(beta+alpha))/(5alphabeta+2[(alpha+beta)^2-2alphabeta])`

By substituting α + β = 3 and αβ = -2 we get,

`S=(3(3))/(5(-2)+2[(3)^2-2xx-2])`

`S=9/(-10+2(13))`

`S=9/(-10+26)`

`S=9/16`

`P=1/(2alpha+beta)xx1/(2beta+alpha)`

`P=1/((2alpha+beta)(2beta+alpha))`

`P=1/(4alphabeta+2beta^2+2alpha^2+betaalpha)`

`P=1/(5alphabeta+2(alpha^2+beta^2))`

`P=1/(5alphabeta+2[(alpha+beta)^2-2alphabeta])`

By substituting α + β = 3 and αβ = -2 we get,

`P=1/(5(-2)+2[(3)^2-2xx-2])`

`P=1/(10+2[9+4])`

`P=1/(10+2(13))`

`P=1/(-10+26)`

`P=1/16`

Hence ,the required polynomial f(x) is given by

`f(x) = k(x^2 - Sx + P)`

`f(x) = k(x^2-9/16x+1/16)`

Hence, the required equation is `f(x) = k(x^2-9/16x+1/16)` Where *k* is any non zero real number.