If the roots of the equation x^3 - 12x^2 + 39 | vedclass

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(C) Let the roots of the equation $x^3 - 12x^2 + 39x - 28 = 0$ be $a - d$,$a$,and $a + d$.According to the properties of roots of a cubic equation:Sum

Vedclass · Accepted Answer

$3$

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(C) Let the roots of the equation $x^3 - 12x^2 + 39x - 28 = 0$ be $a - d$,$a$,and $a + d$.According to the properties of roots of a cubic equation:Sum of roots: $(a - d) + a + (a + d) = -(-12)/1 = 12$$3a = 12 \Rightarrow a = 4$Product of roots: $(a - d) \cdot a \cdot (a + d) = -(-28)/1 = 28$$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$The common difference is $3$ (taking the magnitude).

If the roots of the equation $x^3 - 12x^2 + 39x - 28 = 0$ are in an arithmetic progression,what is the common difference?

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