If the product of roots of the equation $x^{2} - 3 k x + 2 e^{2 l o g_{e} k} - 1 = 0$ is 7, then its roots will real when

None of these

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k = 3

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Solution

The correct option is A
None of these

Given equation can be written as

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

So the product of roots is $2 k^{2}$ – 1. But the product of roots is 7

Hence $2 k^{2} - 1 = 7 \Rightarrow 2 k^{2} = 8 \Rightarrow k = \pm 2$

But $k$ cannot be negative

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