Number of positive integer solutions (a,b) that satisfy $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ where $a ≤ 8$ and $b ≤ 8$

Question

How many ordered pairs $(a, b)$ of positive integers satisfying $a ≤ 8$ and $b ≤ 8$ are there, such that each of the equations

$x^2 + ax + b = 0$, and $x^2 + bx + a = 0$

has two different real solutions in $x$; note all four real solutions need not be different.

I have attempted to rewrite these two equations and then divide to get $x=-(a+b)$ but I am not getting anywhere algebraically when I substitute this back into the original equation. I'm wondering whether I'm approaching this wrong?

Answer 1

Equation $x^2+ax+b=0$ has two distinct real roots if and only if $a^2>4b$ and equation $x^2+bx+a=0$ has two distinct real roots if and only if $b^2>4a$. So, you are after the pairs $(a,b)\in\{1,2,\ldots,8\}^2$ such that $a^2>4b$ and $b^2>4a$. It's not hard to check that these pairs are $(5, 5)$, $(5, 6)$, $(6, 5)$, $(6, 6)$, $(6, 7)$, $(6, 8)$, $(7, 6)$, $(7, 7)$, $(7, 8)$, $(8, 6)$, $(8, 7)$, and $(8, 8)$.

Answer 2

For a general quadratic of the form $ax^{2}+bx+c=0$, the discriminant is given by $\Delta=b^{2}-4ac$. The quadratic has two distinct real roots if and only if $\Delta>0$.

For the two quadratics given, the respective discriminants are $\Delta_{1}=a^{2}-4b$ and $\Delta_{2}=b^{2}-4a$. Hence our goal is to find all positive integers $(a,b)$ such that both $\Delta_{1}$ and $\Delta_{2}$ are positive.

We may assume without loss of generality that $a\geq b$. Then $$\Delta_{1}=a^{2}-4b>0\iff a^{2}>4b\iff a>2\sqrt{b}$$ (since $a$ is positive). However, this also means that $$\Delta_{2}=b^{2}-4a>b^{2}-8\sqrt{b}>0$$ $$\iff b^{2}>8\sqrt{b}\iff b^{\frac{3}{2}}>8\iff b>\sqrt[3]{8^2}=4$$

We continue by listing out the possible $b$:

$$b=5\implies 4a<b^{2}=25\implies a\leq 6\implies (a,b)=(6,5),(5,5)$$

$$b=6\implies 4a<b^{2}=36\implies a\leq 8\implies (a,b)=(8,6),(7,6),(6,6)$$

$$b=7\implies 4a<b^{2}=49\implies a\leq 8\;\mathrm{(as\;the\;question\;states)}\;\implies (a,b)=(8,7),(7,7)$$

Finally, the only solution for $b=8$ is $(a,b)=(8,8)$.

Hence, our solution set is (remember about that WLOG)

$$(a,b)=(5,5),(5,6),(6,5),(6,6),(6,7),(6,8),(7,7),(7,8),(8,6),(8,7),(8,8)$$