1yes sir but i recognised it later when i had already typed the whole part [3]
but since i had typed i didn't want to erase my work so posted in all [4]
3 Answers
1\alpha +\beta = -\frac{b}{a} and \alpha \beta = \frac{c}{a}
\Delta =\begin{vmatrix} 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{vmatrix}
\Delta =\begin{vmatrix} 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{vmatrix}
\Delta =\begin{vmatrix} 1 & 1 & 1\\ 1 & \alpha & \beta \\ 1& \alpha ^{2} & \beta ^{2} \end{vmatrix}X\begin{vmatrix} 1 & 1 & 1\\ 1 & \alpha & \beta \\ 1& \alpha ^{2} & \beta ^{2} \end{vmatrix}
=\Delta _{1} X \Delta _{1} ..................................(say)
\Delta =\Delta _{1}^{2}
\Delta _{1}=\begin{vmatrix} 1 & 1&1 \\ 1& \alpha &\beta \\ 1& \alpha^{2} & \beta^{2} \end{vmatrix}
applying C_{2}\rightarrow C_{2} - C_{1},C_{3}\rightarrow C_{3}-C_{1},then
\Delta _{1}=\begin{vmatrix} 1 & 0&0 \\ 1& \alpha -1 &\beta -1\\ 1& \alpha^{2}-1 & \beta^{2}-1 \end{vmatrix}
EXPANDING ALONG R1
\Delta _{1}=\begin{vmatrix} \alpha -1 & \beta -1\\ \alpha ^{2}-1& \beta^{2} -1 \end{vmatrix}
\Delta _{1}=(\alpha -1)(\beta -1)\begin{vmatrix} \alpha -1 & \beta -1\\ \alpha +1& \beta+1 \end{vmatrix}
\Delta _{1}=\left\{\alpha \beta -(\alpha +\beta )+1 \right\}(\beta -\alpha )
\Delta _{1}=\left\{\alpha \beta -(\alpha +\beta )+1 \right\}\sqrt{\left\{(\alpha +\beta ^{2})-4\alpha \beta \right\}}
\Delta _{1}=\left(\frac{c}{a}+\frac{b}{a}+1 \right)\sqrt{\left(\frac{b^{2}}{a^{2}}-\frac{4c}{a} \right)}
\Delta _{1}=\frac{(a+b+c)\sqrt{b^{2}-4ac}}{a^{2}}
\Delta _{1}^{2}=\frac{(a+b+c)^{2}({b^{2}-4ac})}{a^{4}}
now since
\Rightarrow \Delta =\frac{(a+b+c)^{2}(b^{2}-4ac)}{a^{4}} which is \geq 0
62Manmay, you have proved it way before just by proving that
\Delta =\Delta _1^2
That the determinant is positive.. There was no need of any further work [1]
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