The difference between the corresponding roots of the equation ${{x}^{2}}+ax+b=0$ and ${{x}^{2}}+bx+a=0$ is same, then
( a ) a + b – 4 = 0
( b ) a – b + 4 = 0
( c ) a + b + 4 = 0
( d ) none of these

Hint: In this question, we are given the two quadratic equations. First, we find out the sum and the product of the two quadratic equations and then we find out the difference between their roots, and after solving it, we are able to get the desirable answer.

Formula Used:
Sum of roots = $\dfrac{-b}{a}$
Product of roots = $\dfrac{c}{a}$

Complete step by step Solution:
We are given the equation ${{x}^{2}}+ax+b=0$ and ${{x}^{2}}+bx+a=0$
Let ${{x}_{1}}$ and ${{x}_{2}}$ be the roots of the equation ${{x}^{2}}+ax+b=0$
Compare the above equation with standard form of quadratic equation $a{{x}^{2}}+bx+c=0$, we get
a = 1, b = a and c = b
Therefore sum of roots ( ${{x}_{1}}$+ ${{x}_{2}}$) = $\dfrac{-b}{a}$ = $\dfrac{-a}{1}$= -a
And the product of roots (${{x}_{1}}$${{x}_{2}}$) = $\dfrac{c}{a}$= $\dfrac{b}{1}$ = b
Similarly let ${{y}_{1}}$ and ${{y}_{2}}$ be the roots of the equation ${{x}^{2}}+bx+a=0$
Compare the equation with the standard form of quadratic equation, we get
a = 1, b = b and c = a
Then the sum of roots (${{y}_{1}}+{{y}_{2}}$) = $\dfrac{-b}{a}$= $\dfrac{-b}{1}$= -b
And the product of roots (${{y}_{1}}{{y}_{2}}$) = $\dfrac{c}{a}$= $\dfrac{a}{1}$ = a
Given that ${{x}_{1}}-{{x}_{2}}={{y}_{1}}-{{y}_{2}}$
${{({{x}_{1}}+{{x}_{2}})}^{2}}-4{{x}_{1}}{{x}_{2}}={{({{y}_{1}}+{{y}_{2}})}^{2}}-4{{y}_{1}}{{y}_{2}}$
${{(-a)}^{2}}-4(b)={{(-b)}^{2}}-4(a)$
${{a}^{2}}-4b={{b}^{2}}-4a$
${{a}^{2}}+4a={{b}^{2}}+4b$
By solving it, we get
(a – b)(a + b + 4) = 0
Therefore a – b = 0 and a + b + 4 = 0

Therefore, the correct option is (c).

Note:We also find out the sum and the product of roots by simply putting the formula if x and y are the roots of any quadratic equation the value of xy will be equal to $\dfrac{ constant\, term}{coefficient\, of\, x^2}$ and sum of the roots that is x + y is equal to $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$ and by solving it we get the desired answer.