If the roots of the equation {x^3} - 12{x^2} | vedclass

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Question

(c) Let $a -d, a, a + d$ be the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$

Then $(a - d) + a + (a + d) = 12$ and $(a - d)\,a\,(a + d) =

Vedclass · Accepted Answer

$ \pm 3$

Vedclass · Answer

(c) Let $a -d, a, a + d$ be the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$

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 $a\,({a^2} - {d^2}) = 28$

==> $16 - {d^2} = 7$

$\Rightarrow d = \pm \,3$.

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

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