Question

Find the value of x . 12x^4-56x^3+89x^2-56x+12=0

Answer 1

Note the symmetric coefficients. x=0 is not a root so start by dividing by x^2. This gives.

12(x^2 + 1/x^2) - 56(x+1/x)+89 = 0

Now subtract 12(x+1/x)^2 from both sides and simplify

- 56(x+1/x)+65 = -12(x+1/x)^2

Let u=x+1/x and you now have the quadratic equation

12u^2 - 56u + 65 = 0

(2u-5)(6u-13) = 0

u=5/2 or u=13/6

x+1/x=5/2 gives x=2 or x=1/2 and

x+1/x = 13/6 gives x=3/2 or x=2/3.

Answer 2

Answers

12(x^2 + 1/x^2) - 56(x+1/x)+89 = 0

Now subtract 12(x+1/x)^2 from both sides and simplify

- 56(x+1/x)+65 = -12(x+1/x)^2

Let u=x+1/x and you now have the quadratic equation

12u^2 - 56u + 65 = 0

(2u-5)(6u-13) = 0

u=5/2 or u=13/6

x+1/x=5/2 gives x=2 or x=1/2 and

x+1/x = 13/6 gives x=3/2 or x=2/3.