1. Let the roots of $x^2+ax+b=0$ be $\alpha,\beta$ and the roots of $x^2+bx+a=0$ be $\gamma,\delta$.

2. “Difference between the corresponding roots is same” means the two equations have roots differing by a common constant $k$. So we may write $\gamma=\alpha+k,\qquad \delta=\beta+k.$

3. Using sum of roots:

   • For $x^2+ax+b=0$, $\alpha+\beta=-a.$
   • For $x^2+bx+a=0$, $\gamma+\delta=-b.$

   Hence, $\gamma+\delta=(\alpha+k)+(\beta+k)=\alpha+\beta+2k=-a+2k.$ Therefore, $-b=-a+2k\implies 2k=a-b.$ So, $k=\frac{a-b}{2}.$

4. Using product of roots:

   • For $x^2+ax+b=0$, $\alpha\beta=b.$
   • For $x^2+bx+a=0$, $\gamma\delta=a.$

   Now, $\gamma\delta=(\alpha+k)(\beta+k)=\alpha\beta+k(\alpha+\beta)+k^2.$ Substitute $\alpha\beta=b$ and $\alpha+\beta=-a$: $a=b-ak+k^2.$

5. Substitute $k=\dfrac{a-b}{2}$: $a=b-a\left(\frac{a-b}{2}\right)+\left(\frac{a-b}{2}\right)^2.$

   Multiply by $4$: $4a=4b-2a(a-b)+(a-b)^2.$

   Expand: $4a=4b-2a^2+2ab+a^2-2ab+b^2$ $4a=4b-a^2+b^2.$

   Rearranging: $a^2+4a=b^2+4b$ $a^2-b^2+4a-4b=0$ $(a-b)(a+b)+4(a-b)=0$ $(a-b)(a+b+4)=0.$

6. Given $a\ne b$, we must have $a+b+4=0.$

7. Checking options:

   • A: $a+b+4=0$ ✓
   • B: $a+b-4=0$ ✗
   • C: $a-b-4=0$ ✗
   • D: $a-b+4=0$ ✗

Therefore, the correct option is A.