If a and b can take values 1, 2, 3, 4. Then the number of the equations

Given that the equation $a x^{2}+b x+1=0$.

For given equation to have real roots, $\operatorname{discriminant}(D) \geq 0$

$\Rightarrow b^{2}-4 a \geq 0$

$\Rightarrow b^{2} \geq 4 a$

$\Rightarrow b \geq 2 \sqrt{a}$

Now, it is given that a and b can take the values of 1, 2, 3 and 4.

The above condition $b \geq 2 \sqrt{a}$ can be satisfied when

i) $b=4$ and $a=1,2,3,4$

ii) $b=3$ and $a=1,2$

iii) $b=2$ and $a=1$

So, there will be a maximum of 7 equations for the values of (a, b) = (1, 4), (2, 4), (3, 4), (4, 4), (1, 3), (2, 3) and (1, 2).

Thus, the correct option is (b).