JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Quadratic Equations and Inequalities

The quadratic $a{x}^2+2\sqrt{2}bx+c$ (with $a>0$) is positive for all real $x$ exactly when its discriminant is negative: $(2\sqrt{2}b{)}^2-4ac<0$, i.e. $2b^2<ac$. For $b=1$ the condition $ac>2$ excludes only the pairs with $ac\le 2$ (3 of the 16), leaving $13$; for $b=2$, $ac>8$ holds for $(3,3),(3,4),(4,3),(4,4)$, i.e. $4$ pairs; for $b=3,4$, $2b^2\ge 18>16\ge ac$, so none qualify. The favourable count is $13+4=17$ out of $4^3=64$, so the probability is $\frac{17}{64}$ with $\gcd (17,64)=1$. Therefore $m+n=17+64=81$.