$2$ Similar Quadratics

The simpler way is to use Viète's relations. So, for the equation x^2+ax+b=0, the sum of the roots is m+n = -a (1) and for the equation x^2-bx+a=0, the sum is m+n+28=b (2). From (1) and (2) => 28-a=b => a+b=28

$Here,\quad m\quad and\quad n\quad be\quad the\quad roots\quad of\quad equation\quad { x }^{ 2 }+ax+b\quad =\quad 0\\ So,\quad m=\frac { -a+\sqrt { { a }^{ 2 }-4b } }{ 2 } \quad =>\quad 2m\quad =\quad -a+\sqrt { { a }^{ 2 }-4b } \\ and\quad n\quad =\quad \frac { -a-\sqrt { { a }^{ 2 }-4b } }{ 2 } \quad =>\quad 2n\quad =\quad -a+\sqrt { { a }^{ 2 }-4b } \\ Hence,\quad 2m\quad +\quad 2n\quad =\quad -2a\quad =>\quad m\quad +\quad n\quad =\quad -a\quad ---------\quad (i)\\ \\ Again,\quad m+14\quad and\quad n+14\quad be\quad the\quad roots\quad of\quad equation\quad { x }^{ 2 }-bx+a\quad =\quad 0\\ So,\quad m+14\quad =\frac { -(-b)+\sqrt { { b }^{ 2 }-4a } }{ 2 } \quad =>\quad 2(m+14)\quad =\quad b+\sqrt { { b }^{ 2 }-4a } \\ and\quad n+14\quad =\frac { -(-b)-\sqrt { { b }^{ 2 }-4a } }{ 2 } \quad =>\quad 2(n+14)\quad =\quad b-\sqrt { { b }^{ 2 }-4a } \\ Hence,\quad 2(m+n+14+14)\quad =\quad 2b\quad =>\quad m+n+28\quad =\quad b\quad \quad ----------(ii)\\ \\ Now,\quad Equation\quad (ii)\quad -\quad Equation\quad (i)\\ b-(-a)\quad =\quad m+n+28-m-n\\ =>\quad a+b=28\\ \\ So,\quad value\quad of\quad a+b\quad =\quad 28\quad (Answer)$ .