Question

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

• A. positive real roots
• B. negative real roots
• C. equal roots
• D. imaginary roots

Answer 1

The correct option is D imaginary roots

$\Delta =\left|\begin{array}{ccc}1+1+1& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\text{[multiplying row by row]}=D^2\text{(say)}$
Now,
$\begin{array}{cc}D& =\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\\ & =(1-\alpha )(\alpha -\beta )(\beta -1)\\ & =(\beta -\alpha )[\alpha \beta -\alpha -\beta +1]\\ & =(\beta -\alpha )(\frac{c}{a}+\frac{b}{a}+1)=\frac{(\beta -\alpha )}{a}(a+b+c)\\ \therefore \Delta & =D^2=\frac{(\beta -\alpha {)}^2}{a^2}(a+b+c{)}^2\\ & =\frac{1}{a^2}(a+b+c{)}^2[\frac{b^2}{a^2}-4\frac{c}{a}]\\ & =\frac{1}{a^4}(a+b+c{)}^2(b^2-4ac)\end{array}$

If $\Delta <0$, i.e., $b^2-4ac<0$, then the roots of $a{x}^2+bx+c=0$ are imaginary.