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

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Solution

The correct option is A
Let $a - d, a, a + d$ be the roots of the equation $x^{3} - 12 x^{2} + 39 x - 28 = 0$
Then $(a - d) + a + (a + d) = 12$ and $(a - d) a (a - d) = 28$
$\Rightarrow 3 a = 12$ and $a (a^{2} - d^{2}) = 28$
$\Rightarrow a = 4$ and $a (a^{2} - d^{2}) = 28$
$\Rightarrow 16 - d^{2} = 7 \Rightarrow d = \pm 3.$

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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

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 find the common difference.