Solve the equation `12^x-56 x^2+89 x^2-56 x+12=0.`

Explanation

The given equation is
`12x^(4) - 56x^(3) + 89x^(2) - 56x + 12 = 0 `
Dividing by `x^(2)`, we get
`12x^(2)- 56+ 89x^(2)- (56)/(x)+ (12)/(x^(2))=0`
or `12(x^(2)+(1)/(x^(2))) - 56 (x+(1)/(x)) +89= 0`
or `12[(x+(1)/(x))^(2) -2] - 56 (x+(1)/(x)) +89= 0`
or `12[(x+(1)/(x))^(2) - 56 (x+(1)/(x)) + 65 = 0`
`rArr 12y^(2) - 56y + 65 = 0, where y = x + (1)/(x)`
or `12y^(2) - 26y - 30y + 65 = 0`
or `(6y - 13)(2y-5) = 0`
or `y = (13)/(6), (5)/(2)`
If ` y = 13//6.` then
` x + (1)/(x)= (13)/(6)`
or `6x^(2) - 13x + 6 = 0`
or `(3x - 2)(2x - 3) = 0`
or ` x = (2)/(3).(3)/(2)`
If `y = 5//2,` then `x + (1)/(x)= (5)/(2)`
or `2x^(2) - 5x + 2 = 0`
or `(x - 2) (2x - 1) = 0`
`x = 2, (1)/(2)`
Hence, the roots of the given equation are `2, 1//2, 2//3, 3//2` .