• # question_answer If the product of roots of the equation${{x}^{2}}-3kx+2{{e}^{2\log k}}-1=0$is 7, then its roots will real when [IIT 1984] A) $k=1$ B) $k=2$ C) $k=3$ D)  None of these

Given equation can be written as ${{x}^{2}}-3kx+2{{k}^{2}}-1=0$ 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.