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Question
If the squared difference of the zeros of the quadratic polynomial f(x) = x2 + px + 45 is equal to 144, find the value of p.
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Solution
Given α and β are the zeros of the quadratic polynomial f(x) = x2 + px + 45
`alpha+beta=-"coefficient of x"/("coefficient of "x^2)`
`=(-p)/1`
= -p
`alphabeta="constant term"/("coefficient of "x^2)`
`=45/1`
= 45
we have,
`(alpha-beta)^2=alpha^2+beta^2-2alphabeta`
`144=(alpha+beta)^2-2alphabeta-2alphabeta`
`144=(alpha+beta)^2-4alphabeta`
Substituting `alpha+beta=-p " and "alphabeta=45`
then we get,
`144=(-p)^2-4xx4`
`144=p^2-4xx45`
`144=p^2-180`
`144+180=p^2`
`324=p^2`
`sqrt(18xx18)=pxxp`
`+-18=p`
Hence, the value of p is ±18.
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