If the roots of the equation $x^{3} - 12 x^{2} + 39 x - 28 = 0$ are in A.P., then find the common difference.

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Solution

Let a - d, a, a + d be the roots of the equation, where d is the common difference

$x^{3} - 12 x^{2} + 39 x - 28 = 0$

Then (a - d) + a + (a + d) = 12 and (a - d)a(a + d) = 28

⇒3a = 12 and $a (a^{2} - d^{2})$ = 28

⇒a = 4 and $a (a^{2} - d^{2})$ = 28

⇒ $16 - d^{2}$ = 7 ⇒ d = ± 3.

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x^{3} - 12 x^{2} + 39 x - 28 = 0

x^{3} - 12 x^{2} + 39 x - 28 = 0

x^{3} - 12 x^{2} + 39 x - 28 = 0

If the roots of the equation $x^{3} - 12 x^{2} + 39 x - 28 = 0$ are in A.P., then their common difference will be

x^{3} - 12 x^{2} + 39 x - 28 = 0

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