JEE Main 2026 — Quadratic Equation Question with Solution

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

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

The discriminant condition is:
$D<0$
$\Rightarrow (2\sqrt{2}b{)}^2-4ac<0$
$\Rightarrow 8b^2-4ac<0$
$\Rightarrow 2b^2<ac$

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

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

Case 1: $b=1$
We need $2(1{)}^2<ac\Rightarrow ac>2$.
The total number of pairs $(a,c)$ is $4\times 4=16$.
The pairs for which $ac\le 2$ are $(1,1),(1,2),(2,1)$, which are $3$ in number.
So, the number of pairs with $ac>2$ is $16-3=13$.

Case 2: $b=2$
We need $2(2{)}^2<ac\Rightarrow ac>8$.
The possible pairs $(a,c)$ from the given set are $(3,3),(3,4),(4,3),(4,4)$.
So, there are $4$ pairs.

Case 3: $b=3$
We need $2(3{)}^2<ac\Rightarrow ac>18$.
Since the maximum possible value of $ac$ is $4\times 4=16$, there are $0$ pairs.

Case 4: $b=4$
We need $2(4{)}^2<ac\Rightarrow ac>32$.
Again, there are $0$ pairs.

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

The required probability is $\frac{m}{n}=\frac{17}{64}$.

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

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

Answer: $81$