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

User Keeper
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1 Answer

10 votes
10 votes

Answer:

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

User One Crayon
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