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प्रश्न
If α and β are the zeros of the quadratic polynomial p(s) = 3s2 − 6s + 4, find the value of `alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta`
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उत्तर
Since α and β are the zeros of the quadratic polynomial p(s) = 3s2 − 6s + 4
`alpha+beta="-coefficient of x"/("coefficient of "x^2)`
`alpha+beta=(-(-6))/3`
`alpha+beta=6/3`
`alpha+beta=2`
`alphabeta="constant term"/("coefficient of "x^2)`
`alphabeta=4/3`
We have, `alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta`
`=(alpha^2+beta^2)/(alphabeta)+2[1/alpha+1/beta+3alphabeta]`
`=((alpha+beta)^2-2alphabeta)/(alphabeta)+2[(alpha+beta)/(alphabeta)]+3alphabeta`
By substituting `alpha+beta=2 " and "alphabeta=4/3` we get,
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=((2)^2-2(4/3))/(4/3)+2((2))/(4/3)+3(4/3)`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=(4-8/3)/(4/3)+4/(4/3)+12/3`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=((4xx3)/(1xx3)-8/3)/(4/3)+4/(4/3)+12/3`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=((12-8)/3)/(4/3)+4/(4/3)+12/3`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=(4/3)/(4/3)+4/(4/3)+12/3`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=4/3xx3/4+(4xx3)/4+12/3`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=1+12/4+12/3`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=(1xx12)/(1xx12)+(12xx3)/(4xx3)+(12xx4)/(3xx4)`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=(12+36+48)/12`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=(48+48)/12`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=96/12`
`alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta=8`
Hence, the value of `alpha/beta+beta/alpha+2[1/alpha+1/beta]+3alphabeta " is "8`
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