Question
In the quadratic equation $$a{x^2} + bx + c = 0,\Delta = {b^2} - 4ac{\text{ and }}\alpha {\text{ + }}\beta {\text{,}}{\alpha ^2} + {\beta ^2},{\alpha ^3} + {\beta ^3},$$ are in G.P. where $$\alpha ,\beta $$ are the root of $$a{x^2} + bx + c = 0,$$ then
A.
$$\Delta \ne 0$$
B.
$$b\Delta = 0$$
C.
$$c\Delta = 0$$
D.
$$\Delta = 0$$
Answer :
$$c\Delta = 0$$
Solution :
$$\eqalign{
& {\text{In the quadratic equation }}a{x^2} + bx + c = 0 \cr
& \Delta = {b^2} - 4ac{\text{ and }}\alpha + \beta = - \frac{b}{a},\alpha \beta = \frac{c}{a} \cr
& {\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta \cr
& = \frac{{{b^2}}}{{{a^2}}} - \frac{{2c}}{a} = \frac{{{b^2} - 2ac}}{{{a^2}}} \cr
& {\text{and }}\,\,{\alpha ^3} + {\beta ^3} = - \frac{{{b^3}}}{{{a^3}}} - \frac{{3c}}{a}\left( { - \frac{b}{a}} \right) \cr
& = - \left( {\frac{{{b^3} - 3abc}}{{{a^3}}}} \right) \cr
& {\text{Given }}\alpha + \beta ,{\alpha ^2} + {\beta ^2},{\alpha ^3} + {\beta ^3}{\text{ are in G}}{\text{.P}}{\text{.}} \cr
& \Rightarrow \,\, - \frac{b}{a}, - \frac{{{b^2} - 2ac}}{{{a^2}}}, - \frac{{\left( {{b^3} - 3\,abc} \right)}}{{{a^3}}}{\text{are in G}}{\text{.P}}{\text{.}} \cr
& \Rightarrow {\left( {\frac{{{b^2} - 2\,ac}}{{{a^2}}}} \right)^2} = \frac{b}{a}\left( {\frac{{{b^3} - 3\,abc}}{{{a^3}}}} \right) \cr
& \Rightarrow \,{b^4} + 4{a^2}{c^2} - 4a{b^2}c = {b^4} - 3\,a{b^2}c \cr
& \Rightarrow \,4{a^2}{c^2} - a{b^2}c = 0 \cr
& \Rightarrow \,ac\,\Delta = 0 \cr
& \Rightarrow \,\,c\,\Delta = 0\,\,\left( {\because \,{\text{In quadractic }}a \ne 0} \right) \cr} $$