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

- 1, 2, 5, 8

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- 1, 1, 3, 5

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1, 2, 3, 4

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1, 3, 5, 7

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Solution

The correct option is A $- 1, 1, 3, 5$
As roots are in A.P then let $(a - 3 d), (a - d), (a + d), (a + 3 d)$ be the roots of $x^{4} - 8 x^{3} + 14 x^{2} + 8 x - 15 = 0$
$s_{1} = 8 \Rightarrow a = 2 s_{4} = - 15 \Rightarrow (a^{2} - d^{2}) (a^{2} - 9 d^{2}) = 15 \Rightarrow (4^{2} - d^{2}) (4^{2} - 9 d^{2}) = - 15 \Rightarrow d^{2} = \frac{31}{2}, 1 \Rightarrow d = \pm \sqrt{\frac{31}{2}}, \pm 1$
For $d = \pm 1$
A.P is $- 1, + 1, 3, 5$

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