AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Consider the equation.
6x6−25x5+31x4−31x2+25x−6=0
Simplify the equation as follows,
6(x6−1)−25x(x4−1)+31x2(x2−1)=0
(x2−1)[6(x4+x2+1)−25(x2+1)+31x2]=0
(x2−1)[6x4−25x3+37x2−25x+6]=0
So
x=±1
Or
6x4−25x3+37x2−25x+6=0
6(x4+1)−25x(x2+1)+37x2=0
6(x+‌‌

1

x

)2−25(x+‌‌

1

x

)+37=0
3(x+‌‌

1

x

)[2(x+‌‌

1

x

)−5]−5[2(x+‌‌

1

x

)−5]=0
Solve further
[2(x+‌‌

1

x

)−5][3(x+‌‌

1

x

)−5]=0
Thus
2(x+‌‌

1

x

)−5=0
Or
3(x+‌‌

1

x

)−5
Therefore,
2x2−5x+2=0
2x(x−2)−1(x−2)=0
(x−2)(2x−1)=0
x=2,‌‌

1

2

And
3x2−5x+3=0
It has no solution because D<0
The sum of all rational roots is,
Sum=1−1+2+‌‌

1

2

=2.5