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The equation whose roots are greater by 1 than those of $2 x^{2} - 3 x + 1 = 0$ is

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$3 x^{2} + 5 x - 7 = 0$

$2 x^{2} - 7 x + 6 = 0$

$3 x^{2} - 5 x - 2 = 0$

$2 x^{2} + 5 x + 7 = 0$

answer is B.

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Detailed Solution

Let $α + 1 = x$

$\Rightarrow α = x - 1$

$\Rightarrow f (α) = 0$

$\Rightarrow f (x - 1) = 0$

$2 {(x - 1)}^{2} - 3 (x - 1) + 1 = 0$

$2 (x^{2} + 1 - 2 x) - 3 x + 3 + 1 = 0$

$2 x^{2} + 2 - 4 x - 3 x + 4 = 0$

$2 x^{2} - 7 x + 6 = 0$

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