Question

If $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$ then match the elements of list I with elements of list II:

List I	List II
A) $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=$	1) $\frac{c^2}{a^2}$
B) $\frac{{\alpha }^2+{\beta }^2}{{\alpha }^{-2}+{\beta }^{-2}}=$	2) $\frac{c^5[3abc-b^3]}{a^8}$
C) ${\alpha }^3+{\beta }^3$	3) $\frac{b^2-2ac}{ac}$
D) ${\alpha }^5{\beta }^8+{\alpha }^8{\beta }^5=$	4) $\frac{3abc-b^3}{a^3}$

The correct match from List-I to List-II: $A,B,C,D$

• A. $3,1,4,2$
• B. $1,2,5,4$
• C. $2,3,4,5$
• D. $1,2,3,4$

Answer 1

The correct option is A $3,1,4,2$

As $\alpha ,\beta$ are roots of $a{x}^2+bx+c=0,$ then

$S_1=\alpha +\beta =-\frac{b}{a}$

$S_2=\alpha \beta =\frac{c}{a}$

(A) $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{{(\alpha +\beta )}^2-2\alpha \beta }{\alpha \beta }$

$=\frac{{\left(\frac{-b}{a}\right)}^2-2\left(\frac{c}{a}\right)}{\left(\frac{c}{a}\right)}=\frac{\frac{b^2}{a^2}-\frac{2c}{a}}{\frac{c}{a}}=\frac{b^2-2ac}{ca}$

(B) $\frac{{\alpha }^2+{\beta }^2}{ca}=\frac{{\alpha }^2+{\beta }^2}{\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}}$

$=\frac{{\alpha }^2+{\beta }^2}{\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}}={\alpha }^2+{\beta }^2={\left(\frac{c}{a}\right)}^2=\frac{c^2}{a^2}$

(C) ${\alpha }^2+{\beta }^2={(\alpha +\beta )}^3-3\alpha \beta (\alpha +\beta )={\left(\frac{-b}{a}\right)}^3-3\left(\frac{c}{a}\right)\left(\frac{-b}{a}\right)$

$=\frac{-b^3}{a^3}+\frac{3bc}{a^2}=\frac{-b^3+3abc}{a^3}$

(D) ${\alpha }^5{\beta }^8+{\alpha }^8{\beta }^5={\alpha }^5{\beta }^5({\beta }^3+{\alpha }^3)={\left(\frac{c}{a}\right)}^5\left(\frac{3abc-b^3}{a^3}\right)$

$=\frac{c^5(3abc-b^3)}{a^8}$