We start with the equation:
6x4−5x3+13x2−5x+6=0.
Divide both sides by
x2 (where
x≠0 ) to get:
6x2−5x+13−‌+‌=0Rewrite the expression by grouping terms:
6(x2+‌)−5(x+‌)+13=0Let
t=x+‌. Then
x2+‌=t2−2.
Substitute these into the equation:
6(t2−2)−5t+13=0Simplify:
6t2−12−5t+13=06t2−5t+1=0This 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=‌=‌‌ or ‌‌t=‌‌‌‌ or ‌‌‌‌ So,
x+‌=‌ or
x+‌=‌.
Now solve for
x in each case:
If
x+‌=‌, multiply both sides by
x:x2+1=‌x. Rearranged:
2x2−x+2=0If
x+‌=‌, multiply both sides by
x:x2+1=‌x. Rearranged:
3x2−x+3=0For 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.