Question

solve : 3y^4-13y^2+12=0​

Answer 1

Answer:

Let y^2 = t

so the equation becomes,

3t^2 - 13t +12 = 0

3t^2 - 9t - 4t +12 = 0 (middle term splitting)

3t (t - 3) -4 (t-3) = 0

(t-3) (3t-4) = 0

so, t -3 = 0 => t = 3

and 3t -4 = 0 => t = 4/3

Now, t = y^2 (we let this in the starting)

so, y^2 = 3 or 4/3

thus, y = √3 and √4/3 or 2/√3

Hence the 2 values of y are : √3 and 2/√3

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