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Question
If the zeros of the polynomial f(x) = 2x3 − 15x2 + 37x − 30 are in A.P., find them.
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Solution
Let α = a - d, β = a and γ = a + d be the zeros of the polynomial
f(x) = 2x3 − 15x2 + 37x − 30
Therefore
`alpha+beta+gamma=("coefficient of "x^2)/("coefficient of "x^3)`
`=-((-15)/2)`
`=15/2`
`alphabetagamma="-constant term"/("coefficient of "x^2)`
`=-((-30)/2)`
= 15
Sum of the zeroes `=("coefficient of "x^2)/("coefficient of "x^3)`
`(a-d)+a+(a+d)=15/2`
`a+a+a-d+d=15/2`
`3a=15/2`
`a=15/2xx1/3`
`a=5/2`
Product of the zeroes `="-constant term"/("coefficient of "x^2)`
`alphabetagamma=15`
`(a-d)+a+(a+d)=15`
`a(a^2-d^2)=15`
Substituting a = 5/2 we get
`5/2((5/2)^2-d^2)=15`
`5/2(25/4-d^2)=15`
`25/4-d^2=15xx2/5`
`25/4-d^2=3xx2`
`25/4-d^2=6`
`-d^2=6-25/4`
`-d^2=(24-25)/4`
`-d^2=(-1)/4`
`d^2=1/4`
`d xx d=1/2xx1/2`
`d=1/2`
Therefore, substituting a=5/2 and d=1/2 in α = a - d, β = a and γ = a + d
α = a - d
`alpha=5/2-1/2`
`alpha=(5-1)/2`
`alpha=4/2`
`alpha=2`
β = a
`beta=5/2`
γ = a + d
`gamma=5/2+1/2`
`gamma=(5+1)/2`
`gamma=6/2`
`gamma=3`
Hence, the zeros of the polynomial are 2, `5/2` , 3.
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