3 years ago

X^4+2x^2-48=0 what is the solution set

Mathematics

1 answer:

jolli1 [7]3 years ago

6 0

Answer:

x. = sqrt 6 and x = - sqrt 6

Step-by-step explanation:

x^4 + 2x^2-48=0

(x^2 - 6) (x^2 + 8) = 0

x^2 - 6 = 0

x = + sqrt 6 and x = - sqrt 6

x^2 + 8 = 0

x^2 = sqrt -8 (a negative number doesn’t have a real square root)

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