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

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Question

(C) Let the roots of the cubic equation ${x^3} - 12{x^2} + 39x - 28 = 0$ be $a - d, a, a + d$,which are in $A.P.$According to the properties of roots

Vedclass · Accepted Answer

$\pm 3$

Vedclass · Answer

(C) Let the roots of the cubic equation ${x^3} - 12{x^2} + 39x - 28 = 0$ be $a - d, a, a + d$,which are in $A.P.$According to the properties of roots of a cubic equation:Sum of roots = $(a - d) + a + (a + d) = -(-12)/1 = 12$$3a = 12 \implies a = 4$Product of roots = $(a - d) \cdot a \cdot (a + d) = -(-28)/1 = 28$$a(a^2 - d^2) = 28$Substitute $a = 4$ into the product equation:$4(4^2 - d^2) = 28$$16 - d^2 = 7$$d^2 = 9$$d = \pm 3$Thus,the common difference is $\pm 3$.

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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