If one root of the equation ax^2 + bx + c = 0 | vedclass

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(B) Let the roots of the equation $ax^2 + bx + c = 0$ be $\alpha$ and $\alpha^2$.From the relation between roots and coefficients:Sum of roots: $\alph

Vedclass · Accepted Answer

$(a - b)^3$

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(B) Let the roots of the equation $ax^2 + bx + c = 0$ be $\alpha$ and $\alpha^2$.From the relation between roots and coefficients:Sum of roots: $\alpha + \alpha^2 = -\frac{b}{a} \dots (1)$Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = \frac{c}{a} \dots (2)$From $(1)$,$\alpha(1 + \alpha) = -\frac{b}{a}$. Cubing both sides:$\alpha^3(1 + \alpha)^3 = -\frac{b^3}{a^3}$Substituting $\alpha^3 = \frac{c}{a}$ and $(1 + \alpha)^3 = 1 + \alpha^3 + 3\alpha(1 + \alpha)$:$\frac{c}{a} [1 + \frac{c}{a} + 3(-\frac{b}{a})] = -\frac{b^3}{a^3}$$\frac{c}{a} [\frac{a + c - 3b}{a}] = -\frac{b^3}{a^3}$$c(a + c - 3b) = -b^3$$ac + c^2 - 3bc = -b^3$$b^3 + c^2a + ca^2 = 3abc$This identity can be rearranged as $a(c - b)^3 = c(a - b)^3$.Thus,$X = (a - b)^3$.

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

If one root of the equation $ax^2 + bx + c = 0$ is the square of the other,then $a(c - b)^3 = cX$,where $X$ is

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