Mathematics - Quadratic Equation Question with Solution | TestHub

The quadratic equation $p{x}^2+qx+1=0$ has real roots if its discriminant is greater than or equal to zero. That is, $q^2-4p\ge 0$, or $q^2\ge 4p$.

Given that $p,q\in \{1,2,3,4\}$, we need to find the pairs $(p,q)$ that satisfy this condition. The possible pairs are:

• If $p=1$: $q^2\ge 4⟹q\in \{2,3,4\}$. Pairs: $(1,2),(1,3),(1,4)$.

• If $p=2$: $q^2\ge 8⟹q\in \{3,4\}$. Pairs: $(2,3),(2,4)$.

• If $p=3$: $q^2\ge 12⟹q\in \{4\}$. Pair: $(3,4)$.

• If $p=4$: $q^2\ge 16⟹q\in \{4\}$. Pair: $(4,4)$.

Thus, the required pairs $(p,q)$ are $(1,2),(1,3),(1,4),(2,3),(2,4),(3,4),(4,4)$. The total number of such pairs is 7.