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        Let a and b be the roots of equation Ax^2+Bx+C=0.  Evaluate the value of determinant | 3                      1+a+b              1+a^2 +b^2   || 1+a+b.            1+a^2+b^2      1+a^3+b^3    || 1+a^2+b^2.   1+a^3+b^3.     1+a^4+b^4.   |
2 years ago

## Answers : (1)

mycroft holmes
272 Points
							The given determinant can be written as a product:$\begin{vmatrix} 1 & 1 &1 \\ 1 & a& b\\ 1& a^2& b^2 \end{vmatrix} \ \times \begin{vmatrix} 1 & 1 &1 \\ 1 & a& b\\ 1& a^2 & b^2 \end{vmatrix}$ = $\begin{vmatrix} 1 & 1 &1 \\ 1 & a& b\\ 1& a^2& b^2 \end{vmatrix} ^2$ Using the well known expansion for the Vandermonde Determinant, we can evaluate this as $(a-b)^2 (a-1)^2(b-1)^2$ Since a,b are roots of the quadratic Ax2+Bx+C, we have $(a-b)^2 = \frac{B^2-4AC}{A^2}$ and $(a-1)^2 (b-1)^2 = [(a-1)(b-1)]^2 = P(1)^2 = (A+B+C)^2$ Hence given expressions equals $\frac{(B^2-4AC)(A+B+C)^2}{A^2}$

2 years ago
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• Mind Map
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• NCERT Solutions
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• Previous Year Exam Questions