Let alpha and beta be the roots of the quadra | vedclass

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(D) The condition for the roots of $ax^2 + bx + c = 0$ to be in the ratio $m:n$ is $mnb^2 = ac(m+n)^2$.$(i)$ If $\alpha = \beta$,then the ratio is $1:

Vedclass · Accepted Answer

$i-E, ii-B, iii-D, iv-A$

Vedclass · Answer

(D) The condition for the roots of $ax^2 + bx + c = 0$ to be in the ratio $m:n$ is $mnb^2 = ac(m+n)^2$.$(i)$ If $\alpha = \beta$,then the ratio is $1:1$. Substituting $m=1, n=1$ into the formula: $(1)(1)b^2 = ac(1+1)^2 \Rightarrow b^2 = 4ac$. This matches $(E)$.$(ii)$ If $\alpha = 2\beta$,then the ratio is $2:1$. Substituting $m=2, n=1$: $(2)(1)b^2 = ac(2+1)^2 \Rightarrow 2b^2 = 9ac$. This matches $(B)$.$(iii)$ If $\alpha = 3\beta$,then the ratio is $3:1$. Substituting $m=3, n=1$: $(3)(1)b^2 = ac(3+1)^2 \Rightarrow 3b^2 = 16ac$. This matches $(D)$.$(iv)$ If $\alpha = \beta^2$,then $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$. Substituting $\alpha = \beta^2$,we get $\beta^2 + \beta = -b/a$ and $\beta^3 = c/a$. Thus $\beta = (c/a)^{1/3}$. Substituting this into $\beta^2 + \beta = -b/a$: $(c/a)^{2/3} + (c/a)^{1/3} = -b/a$. Multiplying by $a$: $a(c/a)^{2/3} + a(c/a)^{1/3} = -b \Rightarrow (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$. This matches $(A)$.Therefore,the correct matching is $(i)-E, (ii)-B, (iii)-D, (iv)-A$.

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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