Question

If If α and β are the zeros of the quadratic polynomial f(x) = x² – 2x + 3, find a polynomial whose roots are `(alpha-1)/(alpha+1)` , `(beta-1)/(beta+1)`

Answer 1

Since α and β are the zeros of the quadratic polynomial f(x) = x² – 2x + 3

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

`=(-(-2))/1`

= 2

Product of the zeroes `="constant term"/("coefficient of "x^2)`

`=3/1`

= 3

Let S and P denote respectively the sums and product of the polynomial whose zeros

`(alpha-1)/(alpha+1)` , `(beta-1)/(beta+1)`

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

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

`S=(alphabeta-beta+alpha-1+alphabeta-alpha+beta-1)/(alphabeta+beta+alpha+1)`

`S=(alphabeta+alphabeta-1-1)/(alphabeta+(alpha+beta)+1)`

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

`S=(3+3-1-1)/(3+2+1)`

`S=(6-2)/6`

`S=4/6`

 

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

`P=(alphabeta-beta-alpha+1)/(alphabeta+beta+alpha+1)`

`P=(alphabeta-(beta+alpha)+1)/(alphabeta+(alpha+beta)+1)`

`P=(3-2+1)/(3+2+1)`

`P=2/6`

`P=1/3`

The required polynomial f (x) is given by,

f(x) = k(x² - Sx + P)

`f(x) = k(x^2 - 2/3x + 1/3)`

Hence, the required equation is `f(x) = k(x^2 - 2/3x + 1/3)` , where k is any non zero real number .