Question

If $\alpha ,\beta$ be the roots of the equation $a{x}^2+bx+c=0$. Let $S_n={\alpha }^n+{\beta }^n,$ for $n\ge 1$
If $\Delta =\left|\begin{array}{ccc}3& 1+S_1& 1+S_2\\ 1+S_1& 1+S_2& 1+S_3\\ 1+S_2& 1+S_3& 1+S_4\end{array}\right|$, then $\Delta$ is equal to

• A. $\frac{s^2(b^2-4ac)}{a^4}$
• B. $\frac{(a+b+c{)}^2(b^2-4ac)}{a^4}$
• C. $\frac{b^2-4ac}{a^4}$
• D. $\frac{(a+b+c{)}^2}{4}$

Answer 1

The correct option is B $\frac{(a+b+c{)}^2(b^2-4ac)}{a^4}$

We have $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$, then $\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$
$\therefore S_n={\alpha }^n+{\beta }^n$
$\therefore \Delta =\left|\begin{array}{ccc}3& 1+S_1& 1+S_2\\ 1+S_1& 1+S_2& 1+S_3\\ 1+S_2& 1+S_3& 1+S_4\end{array}\right|$
$=\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|$
$={\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|}^2$
$={\{\alpha \beta -(\alpha +\beta )+1\}}^2\left\{\right(\alpha +\beta {)}^2-4\alpha \beta \}={(\frac{c}{a}+\frac{b}{a}+1)}^2(\frac{b^2}{a^2}-\frac{4c}{a})=\frac{(a+b+c{)}^2(b^2-4ac)}{a^4}$