If each root of the equation $2 x^{3} + {ax}^{2} - 8 x + b = 0$ is reduced by one, then in the transformed equation thus formed, the term containing $x^{2}$ and the constant term are vanishing. The roots of the original equation are

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1,-3,2

$1, 1 \pm \sqrt{7}$

1,1, 6

$1, 3 \sqrt{2,} - \sqrt{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} L e t f (x) = 2 x^{3} + a x^{2} - 8 x + b = 0 \\ I f t h e r o o t s a r e r e d u c e d b y o n e t h e n t h e t r a n s f o r m e d E q u a t i o n i s f (x + 1) = 0 \\ \Rightarrow 2 {(x + 1)}^{3} + a {(x + 1)}^{2} - 8 (x + 1) + b = 0 \\ \Rightarrow 2 (x^{3} + 1 + 3 x^{2} + 3 x) - a (x^{2} + 1 + 2 x) - 8 x - 8 + b = 0 \\ \Rightarrow 2 x^{3} + 2 + 6 x^{2} + 6 x + a x^{2} + a + 2 a x - 8 x - 8 + b = 0 \\ \Rightarrow 2 x^{3} + (6 + a) x^{2} + (2 a - 2) x + a + b - 6 = 0 \end{array}$

$\begin{array}{l} x^{2} c o . e f f = 0 \Rightarrow 6 + a = 0 \Rightarrow a = - 6 \\ a n d c o n s \tan t t e r m = 0 \Rightarrow a + b - 6 = 0 \Rightarrow b = 12 \\ T h e o r i g i n a l E q u a t i o n i s 2 x^{3} - 6 x^{2} - 8 x + 12 = 0 \\ B y t r a i l a n d e r r o r, x = 1 i s a r o o t \end{array}$

$\begin{array}{l} ∴ (x - 1) (2 x^{2} - 4 x - 12) = 0 \\ \Rightarrow (x - 1) (x^{2} - 2 x - 6) = 0 \\ T h e r o o t s o f x^{2} - 2 x - 6 = 0 a r e \\ x = \frac{2 \pm \sqrt{4 + 24}}{2} \Rightarrow \frac{2 \pm 2 \sqrt{7}}{2} \Rightarrow 1 \pm \sqrt{7} \\ ∴ T h e r o o t s a r e 1, 1 \pm \sqrt{7} \end{array}$