Question

If $\alpha$ and $\beta$are the roots of the quadratic equation$a{x}^2+bx+c=0$, observe the lists given below

List I	List II
A. $\alpha =\beta$	1.${\left(a{c}^2\right)}^{1/3}+{(a}^{}$
B. $\alpha =2\beta$	2. $2b^2=9ac$
C. $\alpha =3\beta$	3.$b^2=6ac$
D. $\alpha ={\beta }^2$	4.$3b^2=16ac$
	5.$b^2=4ac$
	6.${\left(a{c}^2\right)}^{1/3}+{\left(a^{2}c\right)}^{1/3}=b$

• A. $A=5,B=2,C=4,D=6$
• B. $A=5,B=2,C=1,D=4$
• C. $A=5,B=4,C=2,D=6$
• D. $A=5,B=2,C=4,D=1$

Answer 1

The correct option is D

$A=5,B=2,C=4,D=1$

Explanation of the correct option :

Given that $\alpha$ and $\beta$are the roots of the quadratic equation$a{x}^2+bx+c=0$.

We know if the roots of $a{x}^2+bx+c=0$ has in the ratio $m:n$ then $Mn{b}^2=ac{(M+n)}^2$

$$\begin{gathered} A.M=n=1 \\ \Rightarrow b^2=4ac \\ B.M=2,n=1 \\ \Rightarrow 2b^2=ac{\left(3\right)}^2 \\ \Rightarrow 2b^2=9ac \\ C.M=3,n=1 \\ \Rightarrow 3b^2=ac{(3+1)}^2 \\ \Rightarrow 3b^2=16ac \end{gathered}$$

If the roots are$ɑ={\beta }^2$, then${\left(a^{2}c\right)}^{1/(2+1)}+{\left(a{c}^2\right)}^{1/(2+1)}=a–b$

${\left(a^{2}c\right)}^{1/3}+{\left(a{c}^2\right)}^{1/3}=–b$

Hence ,the correct option is D.