If the roots of the equation 32x^3 - 48x^2 + | vedclass

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(B) Let the roots of the cubic equation $32x^3 - 48x^2 + 22x - 3 = 0$ be $a-d$,$a$,and $a+d$.From the sum of the roots,$(a-d) + a + (a+d) = -(\frac{-4

Vedclass · Accepted Answer

$\frac{1}{16}$

Vedclass · Answer

(B) Let the roots of the cubic equation $32x^3 - 48x^2 + 22x - 3 = 0$ be $a-d$,$a$,and $a+d$.From the sum of the roots,$(a-d) + a + (a+d) = -(\frac{-48}{32}) = \frac{48}{32} = \frac{3}{2}$.Thus,$3a = \frac{3}{2}$,which implies $a = \frac{1}{2}$.Since $a = \frac{1}{2}$ is a root,it must satisfy the equation: $32(\frac{1}{2})^3 - 48(\frac{1}{2})^2 + 22(\frac{1}{2}) - 3 = 32(\frac{1}{8}) - 48(\frac{1}{4}) + 11 - 3 = 4 - 12 + 11 - 3 = 0$.Now,use the product of the roots taken two at a time: $(a-d)a + a(a+d) + (a-d)(a+d) = \frac{22}{32} = \frac{11}{16}$.Substituting $a = \frac{1}{2}$: $(\frac{1}{2}-d)(\frac{1}{2}) + \frac{1}{2}(\frac{1}{2}+d) + (\frac{1}{4}-d^2) = \frac{11}{16}$.$\frac{1}{4} - \frac{d}{2} + \frac{1}{4} + \frac{d}{2} + \frac{1}{4} - d^2 = \frac{11}{16}$.$\frac{3}{4} - d^2 = \frac{11}{16}$.$d^2 = \frac{3}{4} - \frac{11}{16} = \frac{12-11}{16} = \frac{1}{16}$.Therefore,the square of the common difference is $\frac{1}{16}$.

If the roots of the equation $32x^3 - 48x^2 + 22x - 3 = 0$ are in arithmetic progression,then the square of the common difference of the roots is

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