Question

if the zeroes of the polynomial f(x) = x^3 - 12x^2 + 39x + k are in AP find the value of k

Answer 1

Solution : Let zeros of polynomial be a - d, a and a + d (Since they are in A.P).

→ a - d + a + a + d = - (- 12)/1

→ 3a = 12

→ a = 4

_____________________________

→ (a - d)a + a(a + d) + (a - d)(a + d) = 39

→ a² - ad + a² + ad + a² - d² = 39

→ 3a² - d² = 39

→ 3(4)² - 39 = d²

→ d = √(48 - 39)

→ d = ± 3

_____________________________

→ (a - d)a(a + d) = - (+ k)

→ a³ - ad² = - k

Case 1 : When d = + 3

→ 4³ - 4(+ 3)² = - k

→ 64 - 36 = - k

→ k = - 28

Case 2 : When d = - 3

→ 4³ - 4(- 3)² = - k

→ 64 - 36 = - k

→ k = -28

∴ In each case final value of k is - 28.

Answer : Value of k is - 28.

Answer 2

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$\bigstar \: \bf f(x) = x {}^{3} - 12x {}^{2} + 39 + k \\ \\ \\ \bf Let \: a - d , \: a \: and \: a + d \: be \: \bf the \: zeroes \: of \: the \: polynomial. \\ \\ \\ \bf ATQ \\ \\\bigstar \:\sf T he \: sum \: of \: the \: zeroes = 12 \\ \\ \sf 3a = 12 \\ \\\sf a = 4 \\ \\ \bf f(a), \\\\ \sf - a {(x)}^{3} - l {}^{2} {(4)}^{2} + 39(4) + k = 0 \\\\ \sf 64 - 192 + 156 + k = 0 \\\\ \sf - 28 = k \\ \\\\ \bf \clubsuit \: k = \boxed{ \boxed{ \bf - 28}}$