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

Given Determinant:

$\Delta = \begin{vmatrix} 3 & 1 + s_1 & 1 + s_2 \\ 1 + s_1 & 1 + s_2 & 1 + s_3 \\ 1 + s_2 & 1 + s_3 & 1 + s_4 \end{vmatrix}$

Using the identities: $s_1 = \alpha + \beta$, $s_2 = \alpha^2 + \beta^2$, $s_3 = \alpha^3 + \beta^3$, $s_4 = \alpha^4 + \beta^4$, and using $3 = 1 + 1 + 1$, we rewrite the matrix as:

$\Delta = \begin{vmatrix} 1 + 1 + 1 & 1 + \alpha + \beta & 1 + \alpha^2 + \beta^2 \\ 1 + \alpha + \beta & 1 + \alpha^2 + \beta^2 & 1 + \alpha^3 + \beta^3 \\ 1 + \alpha^2 + \beta^2 & 1 + \alpha^3 + \beta^3 & 1 + \alpha^4 + \beta^4 \end{vmatrix}$

This can be expressed as the square of a 3x3 determinant:

$\Delta = \left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{matrix} \right|^2$

This is a Vandermonde determinant, and its value is: 

$\left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{matrix} \right| = (\beta - \alpha)^2$

So,

$\Delta = (\beta - \alpha)^2 \cdot [\alpha\beta(\alpha + \beta) + 1]^2$   … (i)

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Now consider the quadratic equation: $ax^2 + bx + c = 0$

Its roots are $\alpha$ and $\beta$. So,

• $\alpha + \beta = -\frac{b}{a}$
• $\alpha\beta = \frac{c}{a}$

Then,

$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \left(-\frac{b}{a}\right)^2 - 4 \cdot \frac{c}{a} = \frac{b^2 - 4ac}{a^2}$

Also, note from (i):

$\Delta = (\alpha - \beta)^2 \cdot [\alpha\beta(\alpha + \beta) + 1]^2$

Substitute the values:

$\Delta = \left(\frac{b^2 - 4ac}{a^2}\right) \cdot \left[\frac{c}{a} \cdot \left(-\frac{b}{a}\right) + 1\right]^2$

$= \left(\frac{b^2 - 4ac}{a^2}\right) \cdot \left(\frac{-bc + a^2}{a^2}\right)^2 = \frac{1}{a^4}(b^2 - 4ac)(a + b + c)^2$

Thus, comparing with the form $\frac{k(a + b + c)^2}{a^4}$, we get:

$k = b^2 - 4ac$

Final Answer: Option (A): $b^2 - 4ac$