If one root of the equation ax^3+bx+c=0 is tw | vedclass

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(B) Let the roots of the equation $ax^3+bx+c=0$ be $2\alpha, \alpha, \beta$. From the relation between roots and coefficients,the sum of roots is $2\a

Vedclass · Accepted Answer

$36b^3+343ac^2=0$

Vedclass · Answer

(B) Let the roots of the equation $ax^3+bx+c=0$ be $2\alpha, \alpha, \beta$. From the relation between roots and coefficients,the sum of roots is $2\alpha + \alpha + \beta = 0$ (since the coefficient of $x^2$ is $0$),so $\beta = -3\alpha$. The product of roots taken two at a time is $(2\alpha)(\alpha) + (\alpha)(\beta) + (2\alpha)(\beta) = \frac{b}{a}$. Substituting $\beta = -3\alpha$: $2\alpha^2 + \alpha(-3\alpha) + 2\alpha(-3\alpha) = \frac{b}{a}$ $\Rightarrow 2\alpha^2 - 3\alpha^2 - 6\alpha^2 = \frac{b}{a}$ $\Rightarrow -7\alpha^2 = \frac{b}{a}$ $\Rightarrow \alpha^2 = -\frac{b}{7a}$. The product of roots is $(2\alpha)(\alpha)(\beta) = -\frac{c}{a}$ $\Rightarrow 2\alpha^2(-3\alpha) = -\frac{c}{a}$ $\Rightarrow -6\alpha^3 = -\frac{c}{a}$ $\Rightarrow \alpha^3 = \frac{c}{6a}$. Squaring both sides of the product equation: $(\alpha^3)^2 = (\frac{c}{6a})^2 \Rightarrow \alpha^6 = \frac{c^2}{36a^2}$. Cubing both sides of the sum of products equation: $(\alpha^2)^3 = (-\frac{b}{7a})^3 \Rightarrow \alpha^6 = -\frac{b^3}{343a^3}$. Equating the two expressions for $\alpha^6$: $\frac{c^2}{36a^2} = -\frac{b^3}{343a^3}$ $\Rightarrow 343ac^2 = -36b^3$ $\Rightarrow 36b^3 + 343ac^2 = 0$.

List-$I$	List-$II$
$(i) \alpha = \beta$	$(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii) \alpha = 2\beta$	$(B) 2b^2 = 9ac$
$(iii) \alpha = 3\beta$	$(C) b^2 = 6ac$
$(iv) \alpha = \beta^2$	$(D) 3b^2 = 16ac$
	$(E) b^2 = 4ac$
	$(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$

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