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Question
If α and β are the zeros of the quadratic polynomial f(x) = x2 − 3x − 2, find a quadratic polynomial whose zeroes are `1/(2alpha+beta)+1/(2beta+alpha)`
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Solution
Since α and β are the zeros of the quadratic polynomial f(x) = x2 − 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.
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