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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)`
Solution
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 .
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