Quadratic equation

since a and b are roots therefore

$a^{2} + 10 a -7 = 0$

$b^{2} + 10 b -7 = 0$

$=a^{12} + 10 a^{11} -7a^{10} = 0$

and

$= b^{12} + 10 b^{11} -7b^{10} = 0$

adding both

$a^{12} + b^{12} + 10( a^{11} + b^{11}) - 7( a^{10} + b^{10}) = 0$

$= a^{12} + b^{12} - 7( a^{10} + b^{10}) = -2X5( a^{11} + b^{11})$

$\frac{a^{12} + b^{12}) - 7( a^{10} + b^{10})}{2( a^{11} + b^{11})} = -5$

we can see that a+b=-10 and a b=-7 then (a^12+b^12+a b(a^10+b^10))/2(a^11+b^11) =( a^12+b^12+a^11 b+b^11 a)/2(a^11+b^11) =(a^11 a+b^11 b+a^11 b+b^11 a)/2(a^11+b^11) =(a^11(a+b)+b^11(a+b))/2(a^11+b^11) =((a+b)(a^11+b^11))/2(a^11+b^11) =(a+b)/2 =-5

c'mon u can do this in ur head. just substitute 7=ab in the given equation and factor.a^11+b^11 factors out leaving (a+b /2)=-5