JEE Main 2026 — Mathematics Algebra

For the quadratic expression $ax^2 + 2\sqrt{2}bx + c > 0$ax2+22​bx+c>0 for all $x \in \mathbb{R}$x∈R, the leading coefficient must be positive and the discriminant must be negative.

Since $a \in \{1, 2, 3, 4\}$a∈{1,2,3,4}, the condition $a > 0$a>0 is always satisfied.

The discriminant condition is:

$$D < 0$$D<0

$$\Rightarrow (2\sqrt{2}b)^2 - 4ac < 0$$⇒(22​b)2−4ac<0

$$\Rightarrow 8b^2 - 4ac < 0$$⇒8b2−4ac<0

$$\Rightarrow 2b^2 < ac$$⇒2b2<ac

The total number of possible triplets $(a, b, c)$(a,b,c) is $4 \times 4 \times 4 = 64$4×4×4=64.

Now, we find the number of favorable outcomes by checking the possible values of $b \in \{1, 2, 3, 4\}$b∈{1,2,3,4}:

Case 1: $b = 1$b=1

We need $2(1)^2 < ac \Rightarrow ac > 2$2(1)2<ac⇒ac>2.

The total number of pairs $(a, c)$(a,c) is $4 \times 4 = 16$4×4=16.

The pairs for which $ac \le 2$ac≤2 are $(1, 1), (1, 2), (2, 1)$(1,1),(1,2),(2,1), which are $3$3 in number.

So, the number of pairs with $ac > 2$ac>2 is $16 - 3 = 13$16−3=13.

Case 2: $b = 2$b=2

We need $2(2)^2 < ac \Rightarrow ac > 8$2(2)2<ac⇒ac>8.

The possible pairs $(a, c)$(a,c) from the given set are $(3, 3), (3, 4), (4, 3), (4, 4)$(3,3),(3,4),(4,3),(4,4).

So, there are $4$4 pairs.

Case 3: $b = 3$b=3

We need $2(3)^2 < ac \Rightarrow ac > 18$2(3)2<ac⇒ac>18.

Since the maximum possible value of $ac$ac is $4 \times 4 = 16$4×4=16, there are $0$0 pairs.

Case 4: $b = 4$b=4

We need $2(4)^2 < ac \Rightarrow ac > 32$2(4)2<ac⇒ac>32.

Again, there are $0$0 pairs.

Total number of favorable outcomes = $13 + 4 = 17$13+4=17.

The required probability is $\dfrac{m}{n} = \dfrac{17}{64}$nm​=6417​.

Since $\gcd(17, 64) = 1$gcd(17,64)=1, we have $m = 17$m=17 and $n = 64$n=64.

Therefore, $m + n = 17 + 64 = 81$m+n=17+64=81.

Answer: $81$81