We start with the equation: 6x4−5x3+13x2−5x+6=0. Divide both sides by x2 (where x≠0 ) to get: 6x2−5x+13−‌
5
x
+‌
6
x2
=0 Rewrite the expression by grouping terms: 6(x2+‌
1
x2
)−5(x+‌
1
x
)+13=0 Let t=x+‌
1
x
. Then x2+‌
1
x2
=t2−2. Substitute these into the equation: 6(t2−2)−5t+13=0 Simplify: 6t2−12−5t+13=06t2−5t+1=0 This is a quadratic equation in t. The discriminant (D) is: D=(−5)2−4⋅6⋅1=25−24=1 (which is greater than 0 ). This means there are two real solutions for t : t=‌
5±1
12
=‌
6
12
‌ or ‌‌
4
12
t=‌
1
2
‌‌‌ or ‌‌‌‌
1
3
So, x+‌
1
x
=‌
1
2
or x+‌
1
x
=‌
1
3
. Now solve for x in each case: If x+‌
1
x
=‌
1
2
, multiply both sides by x:x2+1=‌
1
2
x. Rearranged: 2x2−x+2=0 If x+‌
1
x
=‌
1
3
, multiply both sides by x:x2+1=‌
1
3
x. Rearranged: 3x2−x+3=0 For both equations, the discriminant is less than 0 . This means there are no real solutions for x. Therefore, all the solutions for x are complex numbers.