If the roots of $x^{4} - 8 x^{3} + 14 x^{2} + 8 x - 15 = 0$ are in A.P. then the roots are

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1, 2, 3, 4

1, 3, 5, 7

–1, 1, 3, 5

1, 1, –3, –5

answer is A.

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Detailed Solution

$\begin{array}{l} L e t a - 3 d, a - d, a + d, a + 3 d a r e r o o t s o f x^{4} - 8 x^{3} + 14 x^{2} + 8 x - 15 = 0 \\ ∴ a - 3 d + a - d + a + d + a + 3 d = 8 \\ \Rightarrow 4 a = 8 \Rightarrow a = 2 \\ ∴ (a - 3 d) (a - d) (a + d) (a + 3 d) = - 15 \\ \Rightarrow (a^{2} - 9 d^{2}) (a^{2} - d^{2}) = - 15 \\ \Rightarrow (4 - 9 d^{2}) (4 - d^{2}) = - 15 \\ \Rightarrow 16 - 4 d^{2} - 36 d^{2} + 9 d^{4} = - 15 \\ \Rightarrow 9 d^{4} - 40 d^{2} + 31 = 0 \\ \Rightarrow 9 d^{4} - 31 d^{2} - 9 d^{2} + 31 = 0 \\ \Rightarrow d^{2} (9 d^{2} - 31) - 1 (9 d^{2} - 31) = 0 \\ \Rightarrow (d^{2} - 1) (9 d^{2} - 31) = 0 \\ \Rightarrow d^{2} = 1, d^{2} = \frac{31}{9} \\ ∴ d = 1 \\ T h e r o o t s a r e 2 - 3, 2 - 1, 2 + 1, 2 + 3 \\ = - 1, 1, 3, 5 \end{array}$