studentfromsg studentfromsg
  • 01-04-2019
  • Mathematics
contestada

If[tex](3x^{2} +5x-2)^{2n-1}[/tex] = 1, find the value of n

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LammettHash
LammettHash LammettHash
  • 01-04-2019

Either [tex]3x^2+5x-2=1[/tex], in which case [tex]n[/tex] can be anything, or [tex]3x^2+5x-2\neq0[/tex] and [tex]2n-1=0[/tex], in which case [tex]n=\dfrac12[/tex].

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altavistard
altavistard altavistard
  • 01-04-2019

Answer:

n = 1/2

Step-by-step explanation:

Assuming that 3x² + 5x - 2 is not zero, and that we can represent 3x² + 5x - 2 with the letter d, then:

d^(2n - 1) = 1.  The exponent (2n - 1) of  d  must be 0 for d^(2n - 1) = 1 to be true.  Then 2n - 1 = 0, and 2n = 1, and n = 1/2.

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