Q. The roots of the equation $ax^{3}+bx^{2}+cx+d=0$, are $\alpha_{1}, \alpha_{2}, \alpha_{3}$ and roots of $g (z) = ax^{3}+\frac{f'' (y) z^{2}}{2!}+\frac{f' (y)}{1!}z+f (y)=0$ are $\beta_{1}, \beta_{2}, \beta_{3}$ then $\alpha_{1}-\beta_{1}$ equals
Complex Numbers and Quadratic Equations
Solution:
Let $f (x) = ax^{3}+bx^{2}+cx+d \, \dots (1)$
$g (z) =az^{3}+\frac{f'' (y)}{2!}z^{2}+\frac{f' (y)}{1!}z+f (y)=0 \, \dots (2)$
Now, $f''(y) = 6ay +2b$
$f' (y) = 3ay^{2}+2by +c$
$f (y) = ay^{3}+by^{2}+cy +d$
Putting in (2), we get
$g (z) = az^{3}+\frac {6ay+2b}{2!} z^{2} + (3ay^{2}+2by+c)z+ay^{3}+'by^{2}+cy+d$
$=az^{3}+(3ay+b)z^{2}+(3ay^{2}+2by +c) z+ay^{3}+by^{2}+cy+d$
$=az^{3}+3ayz^{2}+bz^{2}+3ay^{2}z+2byz+cz+ay^{3}+by^{2}+cy+d$
collecting all a, b. c terms, we get
$=a(z^{3}+y^{3}+3z^{2}y +3zy^{2})+b (z^{2}+y^{2}+2zy)+(cz+cy+d)$
$=a(z+y)^{3}+b (z+y)^{2}+c(z+y)+d$
$g (z) = f (z+y)$, which is obtained by replacing x by $z+y$ in equation $(1)$, say $z+y =x$
$\therefore z=x+y$
Hence, which means the roots of (1) are connected by a quantity y
$\therefore $ If $\alpha_{1}, \alpha_{2}, \alpha_{3}$ are roots of (1) and
$\beta_{1}, \beta_{2}, \beta_{3}$ are roots of (2)
$\alpha_{1}-y = \beta_{1}, \alpha_{2}-y= \beta_{2}, \alpha_{3}-y=B_{3}$
$\Rightarrow \alpha_{1}-\beta_{1}=y, \alpha_{2}-\beta_{2}=y, \alpha_{3}-y=B_{3}$
$\Rightarrow \alpha_{1}-\beta_{1}=y, \alpha_{2}-\beta_{2}=y, \alpha_{3}-\beta_{3}=y$
$\therefore \alpha_{1}-\beta_{1}=\alpha_{2}-\beta_{2}=\alpha_{3}-\beta_{3}=y$
$g (z) = az^{3}+\frac {6ay+2b}{2!} z^{2} + (3ay^{2}+2by+c)z+ay^{3}+'by^{2}+cy+d$
$=az^{3}+(3ay+b)z^{2}+(3ay^{2}+2by +c) z+ay^{3}+by^{2}+cy+d$
$=az^{3}+3ayz^{2}+bz^{2}+3ay^{2}z+2byz+cz+ay^{3}+by^{2}+cy+d$
collecting all a, b. c terms, we get
$=a(z^{3}+y^{3}+3z^{2}y +3zy^{2})+b (z^{2}+y^{2}+2zy)+(cz+cy+d)$
$=a(z+y)^{3}+b (z+y)^{2}+c(z+y)+d$
$g (z) = f (z+y)$, which is obtained by replacing x by $z+y$ in equation $(1)$, say $z+y =x$
$\therefore z=x+y$
Hence, which means the roots of (1) are connected by a quantity y
$\therefore $ If $\alpha_{1}, \alpha_{2}, \alpha_{3}$ are roots of (1) and
$\beta_{1}, \beta_{2}, \beta_{3}$ are roots of (2)
$\alpha_{1}-y = \beta_{1}, \alpha_{2}-y= \beta_{2}, \alpha_{3}-y=B_{3}$
$\Rightarrow \alpha_{1}-\beta_{1}=y, \alpha_{2}-\beta_{2}=y, \alpha_{3}-y=B_{3}$
$\Rightarrow \alpha_{1}-\beta_{1}=y, \alpha_{2}-\beta_{2}=y, \alpha_{3}-\beta_{3}=y$
$\therefore \alpha_{1}-\beta_{1}=\alpha_{2}-\beta_{2}=\alpha_{3}-\beta_{3}=y$