Christmas Streak 24/88: Hard To Be Always Positive

We don't know whether $ax^2+bx+c$ is quadratic. Therefore, we must split the cases.

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$i)~a\neq 0$

Then the discriminant $D=b^2-4ac<0.$ Therefore $b^2<4ac.$

Since $a+c=5,$ $(a,~c)=(1,~4),~(2,~3),~(3,~2),~(4,~1).$ However, the last two cases would yield the same values for $a^2+b^2+c^2$ with the first two cases.

So we're going to look at $(a,~c)=(1,~4),~(2,~3).$

If $a=1,~c=4,$ then $b^2<16.$ Possible values for $b^2$ are $0,~1,~4,~9.$

$\therefore~a^2+b^2+c^2=17,~18,~21,~26.$

If $a=2,~c=3,$ then $b^2<24.$ Possible values for $b^2$ are $0,~1,~4,~9,~16.$

$\therefore~a^2+b^2+c^2=13,~14,~17,~22,~29.$

From above, the possible values are $13,~14,~17,~18,~21,~22,~26,~29.$

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$(ii)~a=0$

Then $bx+c>0,$ but if $b\neq 0,$ this doesn't make sense, because a linear graph (except the constant one) always passes through the $x$ -axis.

So $b=0,$ and since $a+c=5,~c=5.$

$\therefore~a^2+b^2+c^2=25.$

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From above, the sum of all the possible values for $a^2+b^2+c^2$ is $13+14+17+18+21+22+25+26+29=\boxed{185}.$