What substitution should be used to rewrite 16(x3 + 1)2 – 22(x3 + 1) – 3 = 0 as a quadratic equation?

  z = x^3 +1

Step-by-step explanation:

Noting the squared term, it makes sense to substitute for that term:

  z = x^3 +1

gives ...

  16z^2 -22z -3 = 0 . . . . the quadratic you want

Solutions derived from that substitution

Factoring gives ...

  16z^2 -24z +2z -3 = 0

  8z(2z -3) +1(2z -3) = 0

  (8z +1)(2z -3) = 0

  z = -1/8  or  3/2

Then we can find x:

  x^3 +1 = -1/8

  x^3 = -9/8 . . . . . subtract 1

  x = (-1/2)∛9 . . . . . one of the real solutions

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  x^3 +1 = 3/2

  x^3 = 1/2 = 4/8 . . . . . . subtract 1

  x = (1/2)∛4 . . . . . . the other real solution

The complex solutions will be the two complex cube roots of -9/8 and the two complex cube roots of 1/2.

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation:

need more info

step-by-step explanation: