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Question: If a, b, g are different from 1 and are the roots of ax<sup>3</sup> + bx<sup>2</sup> + cx + d = 0 a...

If a, b, g are different from 1 and are the roots of

ax3 + bx2 + cx + d = 0 and (b – g) (g – a) (a – b) =252\frac{25}{2}, then the determinant D = ∣α1−αβ1−βγ1−γαβγα2β2γ2∣\left| \begin{matrix} \frac{\alpha}{1 - \alpha} & \frac{\beta}{1 - \beta} & \frac{\gamma}{1 - \gamma} \\ \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \end{matrix} \right|equals:

A

25d2a\frac{25d}{2a}

B

25da\frac{25d}{a}

C

−25da+b+c+d\frac{- 25d}{a + b + c + d}

D

None of these

Answer

None of these

Explanation

Solution

Taking a, b, g common from C1, C2, C3 respectively, we get

D= abg ∣11−α11−β11−γ111αβγ∣\left| \begin{matrix} \frac{1}{1 - \alpha} & \frac{1}{1 - \beta} & \frac{1}{1 - \gamma} \\ 1 & 1 & 1 \\ \alpha & \beta & \gamma \end{matrix} \right|

= abg ∣11−α11−β−11−α11−γ−11−α100αβ−αγ−α∣\left| \begin{matrix} \frac{1}{1 - \alpha} & \frac{1}{1 - \beta} - \frac{1}{1 - \alpha} & \frac{1}{1 - \gamma} - \frac{1}{1 - \alpha} \\ 1 & 0 & 0 \\ \alpha & \beta - \alpha & \gamma - \alpha \end{matrix} \right|

[using C2 ® C2 – C1 and C3 ® C3 – C1]

= αβγ(−1)(β−α)(γ−α)(1−α)(1−β)(1−γ)\frac { \alpha \beta \gamma ( - 1 ) ( \beta - \alpha ) ( \gamma - \alpha ) } { ( 1 - \alpha ) ( 1 - \beta ) ( 1 - \gamma ) } ∣1−γ1−β11∣\left| \begin{matrix} 1 - \gamma & 1 - \beta \\ 1 & 1 \end{matrix} \right|

= αβγ(α–β)(β−γ)(γ−α)(1−α)(1−β)(1−γ)\frac{\alpha\beta\gamma(\alpha –\beta)(\beta - \gamma)(\gamma - \alpha)}{(1 - \alpha)(1 - \beta)(1 - \gamma)}

As a, b, g are the roots of ax3 + bx2 + cx + d = 0, ax3 + bx2 + cx + d = a(x – a) (x – b) (x – g)

and abg = –d/a

Thus, D = (−d/a)(25/2)(a+b+c+d)/a\frac{( - d/a)(25/2)}{(a + b + c + d)/a}= 25d2(a+b+c+d)\frac{25d}{2(a + b + c + d)}.