Question

Let α,β be the roots of the equation, ax²+bx+c=0.a,b,c are real and s_n=αⁿ+βⁿ and \(\begin{vmatrix}3 &1+s_1  &1+s_2\\1+s_1&1+s_2  &1+s_3\\1+s_2&1+s_3  &1+s_4\end{vmatrix}=\frac{k(a+b+c)^2}{a^4}\) then k=

• A. b²-4ac
• B. b²+4ac
• C. b²+2ac
• D. 4ac-b²

Answer 1

The Correct Option is A

The correct answer is option (A): b²-4ac

\(\begin{vmatrix}3 &1+s_1  &1+s_2\\1+s_1&1+s_2  &1+s_3\\1+s_2&1+s_3  &1+s_4\end{vmatrix}\)=\(\begin{vmatrix} 1+1+1 &1+\alpha+\beta  &1+\alpha^2+\beta^2\\   1+\alpha+\beta&1+\alpha^2+\beta^2  &1+\alpha^3+\beta^3 \\   1+\alpha^2+\beta^2&1+\alpha^3+\beta^3  & 1+\alpha^4+\beta^4 \end{vmatrix}\)

\(=\begin{vmatrix} 1 &1  &1 \\   1&\alpha  & \beta \\   1&\alpha^2  &\beta^2  \end{vmatrix}\begin{vmatrix} 1 &1  &1 \\   1&\alpha  & \beta \\   1&\alpha^2  &\beta^2  \end{vmatrix}\)

\(=[\alpha\beta(\beta-\alpha)-\beta^2+\alpha^2+\beta-\alpha][\alpha\beta(\beta-\alpha)-(\beta^2-\alpha^2)+(\beta-\alpha)]\)

\(=(\beta-\alpha)^2[\alpha\beta(\beta+\alpha)+1]^2\)………..(i)

Now, ax²+bx+c=0

here \(\alpha+\beta=-\frac{b}{a}\) and \(\alpha\beta=\frac{c}{a}\)

Now, \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\)

\(\Rightarrow\)\(\frac{b^2}{a^2}-\frac{4c}{a}\) = \(\frac{b^2-4ac}{a^2}\)

Now, \(\Delta\) = \((\frac{b^2-4ac}{a^2})[\frac{c}{a}-(-\frac{b}{a})+1]\)

\(\Rightarrow (\frac{b^2-4ac}{a^2})(\frac{c+a+b}{a})^2\) = \(\frac{1}{a^4}(b^2-4ac)(a+b+c)^2\)

Now,  \(\frac{k(a+b+c)^2}{a^4} =\frac{1}{a^4}(b^2-4ac)(a+b+c)^2\)

\(\Rightarrow k = (b^2-4ac)\)