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Let, both $a$ and $b$ belong to the set {1,2,3,4}. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots.

for real roots, $a \gt 0$, $b^2-4{a}{c} \ge 0$

Here we have $c=1$, and $a \ge 0$

Now we need to have $b^2-4a \ge 0$

i.e. $(-2\sqrt{a} \ge b) \cup (b \ge 2\sqrt{a})$

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I've changed algebra tag to algebra-precalculus, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Aug 24 '12 at 14:55
up vote 4 down vote accepted

You have basically answered the question yourself. You just need $a>0$ which is automatic and $b^2\geq 4ac=4a$. So for $a=1$, $b$ can be $2,3,4$, for $a=2$, $b$ can be $3,4$ and for $a=3$ or $4$, $b$ can be only $4$. Thus there are $7$ possible polynomials.

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There are of course 16 equations of the form $ax^2+bx+1=0$ with $a,b \in X:= \{1,2,3,4\}$. Which of them have real roots?

As you note, we have to determine for which $a,b\in X$ we have $b^2-4a\ge 0$.

So if $a=1$, we have $b=2,3,4$. If $a=2$, we have $b=3,4$, if $a=3,4$ then we must have $b=4$. Summing up there are 7 equations.

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