The condition that the roots of x^3-b x^2+c x | vedclass

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Question

(A) Given the cubic equation: $x^3-b x^2+c x-d=0$. Let the roots in geometric progression be $\frac{a}{r}, a, ar$. From the relationship between roots

Vedclass · Accepted Answer

$c^3=b^3 d$

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(A) Given the cubic equation: $x^3-b x^2+c x-d=0$. Let the roots in geometric progression be $\frac{a}{r}, a, ar$. From the relationship between roots and coefficients: $1$. Sum of roots: $\frac{a}{r} + a + ar = b \Rightarrow a(\frac{1}{r} + 1 + r) = b$ ... $(i)$ $2$. Sum of roots taken two at a time: $\frac{a}{r} \cdot a + a \cdot ar + ar \cdot \frac{a}{r} = c \Rightarrow a^2(\frac{1}{r} + r + 1) = c$ ... (ii) $3$. Product of roots: $\frac{a}{r} \cdot a \cdot ar = d \Rightarrow a^3 = d$ ... (iii) Dividing (ii) by $(i)$: $\frac{a^2(\frac{1}{r} + 1 + r)}{a(\frac{1}{r} + 1 + r)} = \frac{c}{b} \Rightarrow a = \frac{c}{b}$. Substituting $a = \frac{c}{b}$ into (iii): $(\frac{c}{b})^3 = d$ $\Rightarrow \frac{c^3}{b^3} = d$ $\Rightarrow c^3 = b^3 d$.

The condition that the roots of $x^3-b x^2+c x-d=0$ are in geometric progression is

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