If the product of the roots for the equation $x^{2} - 3 k x + 2 e^{2 l n k} - 1 = 0$ is 7,then the roots of the equation are real for k=?.

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Solution

$x^{2} - 3 k x + 2 e^{2 l n k} - 1 = 0$
$\Rightarrow x^{2} - 3 k r + 2 e^{l n k^{2}} - 1 = 0 \Rightarrow x^{2} - 3 k x + 2 k^{2} - 1 = 0$
Product of roots $= 2 k^{2} - 1 = 7 \Rightarrow k = \pm 2$
But lnk gives restriction of, $k > 0$
$D = 9 k^{2} - 4 (2 k^{2} - 1) = k^{2} + 4 > 0$
$∴ k = 2$ .

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