quadratic equation

[man111]

If [tex]c[/tex] and [tex]d[/tex] are the roots of the equation [tex]x^2-10ax-11b=0[/tex] and

If [tex]a[/tex] and [tex]b[/tex] are the roots of the equation [tex]x^2-10cx-11d=0[/tex] . then [tex]a+b+c+d=[/tex]

[martin123456]

Substituting in the first equation [tex]x=c[/tex] we get [tex]c^2-10ac-11b=0[/tex]. Substituting in the second [tex]x=a[/tex] we get [tex]a^2-10ac-11d=0[/tex]. Substructing these we get [tex](c-a)(c+a)=11(b-d)[/tex](3). Substructing (1) and (2) we get [tex](a-c)+(b-d)=-10(a-c) \Rightarrow b-d=-11(a-c)[/tex]. Substitute in (3) to get [tex](c-a)(c+a)=11^2(c-a)[/tex]. If [tex]a=c[/tex] then from (1) and (2) we get [tex]d=b[/tex], so we have [tex]x^2-10ax-11b=0[/tex] has roots [tex]a[/tex] and ptex]b[/tex] and so [tex]a+b=10a[/tex] and [tex]ab=-11b[/tex], so [tex]a=-11[/tex], so [tex]b=-99[/tex], so [tex]a+b=-110[/tex], so [tex]a+b+c+d=2(a+b)=-220[/tex].
Otherwise, [tex]c+a=121[/tex], so the result would be [tex]1210[/tex]. For this case, please check if it is possible.