Given $l x^{2} - m x + 5 = 0$ does not have two distinct real roots, the minimim value of $5 l + m$

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Solution

Considering equal roots.
$b^{2} = 4 a c$
Or
$m^{2} = 20 l$
Or $l = \frac{m^{2}}{20}$
Hence $5 l + m$
$k = \frac{m^{2}}{4} + m$
$\frac{d k}{d m} = \frac{m}{2} + 1$
$= 0$
Or $m = - 2$
Substituting in k, we get
$= \frac{4}{4} - 2$
$= - 1$

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