Don't make mistakes! #1

We can't use the discriminant yet, since we don't know if the equation is quadratic.

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$(i)\quad a\neq 0$

Now we can use the discriminant.

$D/4=24^2-144a=0$

$\therefore a=4$

Then $4x^2-48x+144=4(x^2-12x+36)=4(x-6)^2=0$ .

$k=6$ .

$\therefore \boxed{a+k=10}$ .

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$(ii)\quad a=0$

Then the equation is linear, and therefore has only one solution.

$-48x+144=0$

$k=3$

$\therefore \boxed{a+k=3}$ .

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From $(i)$ and $(ii)$ , we see that the mininum of $a+k$ is $\boxed{3}$ .

just as Mr. H.M. 유 .