Question

How would I find all real solutions in x^4/3 -10^2/3 +21= 0

Answer 1

Answer:

Step-by-step explanation:

This equation looks complicated.We have to make it easier

let's say x^2/3 = t  and  x^4/3 = t^2

     

      t^2-10t+21=0   [ we can factorize this equation as a (t-3)(t-7) ]

     

      (t-3)(t-7)=0  [ that means , t can be 3 or 7 ]

But don't forget we have to find x not t so,

   t=x^2/3=3     ∛ x^2 = 3       x^2 = 9    x=3 or x= -3

  t=x^2/3=7      ∛x^2 = 7        x^2 = 343    x ~18.5  or x ~ -18.5