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If the roots of the equation $x^{2}-4 x-\log _{3} N=0$ are real, then what is the minimum value of $N ?$
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The correct answer is:
$1 / 6$
Given equation is $x^{2}-4 x-\log _{3} N=0$
Since, roots are real $\therefore b^{2}-4 a c=0 \Rightarrow(4)^{2}-4\left(-\log _{3} N\right) \geq 0$
$\Rightarrow 16 \geq-4 \log _{3} N$
$\Rightarrow 4 \geq-\log _{3} N$
$\Rightarrow 4 \geq \log _{3} N^{-1}$
$\Rightarrow N^{-1} \geq 3^{4} \geq 81$
$\Rightarrow N \geq \frac{1}{81}$
Hence, minimum val ue of $\mathrm{N}$ is $\frac{1}{81}$.
Since, roots are real $\therefore b^{2}-4 a c=0 \Rightarrow(4)^{2}-4\left(-\log _{3} N\right) \geq 0$
$\Rightarrow 16 \geq-4 \log _{3} N$
$\Rightarrow 4 \geq-\log _{3} N$
$\Rightarrow 4 \geq \log _{3} N^{-1}$
$\Rightarrow N^{-1} \geq 3^{4} \geq 81$
$\Rightarrow N \geq \frac{1}{81}$
Hence, minimum val ue of $\mathrm{N}$ is $\frac{1}{81}$.
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