Question

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Answer 1

Answer:

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Answer 2

$\underline{ \underline{\huge{ \bf{question}}}}$

If the zeroes of polynomial

$\mathrm{f(x) = {x}^{3} - 12 {x}^{2} + 39x + k }$

are in AP, find the value of 'k'.

$\underline{\underline{ \huge{ \bf{solution}}}}$

Let,

zeroes of f (x) be

a - d , a , a + d

then,

$\boxed{ \tt{sum \: of \: zeroes \: of \: cubic \: polynomial = \frac{ - coefficient \: of \: {x}^{2} }{coeffiecient \: of \: {x}^{3} } }}$

so,

$\mathrm{a - d + a + a + d = \frac{ - ( - 12)}{1} } \\ \\ \mathrm{3a = 12} \\ \\ \red{\mathrm{a = 4}}$

also,

$\boxed{ \tt{(a - d)(a) + (a - d)(a + d) + (a)(a + d) = \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{3} } }}$

so,

$\mathrm{ {a}^{2} - ad + {a}^{2} - {d}^{2} + {a}^{2} + ad = 39 } \\ \\ \mathrm{3 {a}^{2} - {d}^{2} = 39 } \\ \\ \mathrm{3 {(4)}^{2} - {d}^{2} = 39} \\ \\ \mathrm{48 - {d}^{2} = 39 } \\ \\ \red{ \mathrm{d = 3}}$

Again,

$\boxed{\tt{product \: of \: zeroes = \frac{ - constant \: term}{coefficient \: of \: {x}^{3}}} }$

so,

$\mathrm{(a)(a - d)(a + d) = \frac{ - k}{1} } \\ \\ \mathrm{a( {a}^{2} - {d}^{2} ) = - k}$

putting values

$\mathrm{4( {4}^{2} - {3}^{2} ) = - k } \\ \\ \mathrm{4(16 - 9) = - k} \\ \\ \boxed{ \red{ \bf{k = - 28}}}$