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 #1
avatar+130466 
+5
2 Jun 2014
 #1
avatar+130466 
+5

"Roots" would imply that

(x^3  + 125) / (x +5) = 0

Factoring the top as a sum of cubes, we get

(x + 5)(x^2 - 5x + 25)         So we have

(x + 5)(x^2 - 5x + 25) / (x + 5) = 0    Only the numerator can make this zero

So setting both factors to 0, we have

(x + 5 ) = 0      so x =-5, but we must reject this since it makes the denominator of the original function undefined.

The other two roots are non-real...using the on-site solver, we have

x25×x+25=0{x=(5×3×i5)2x=(5×3×i+5)2}{x=(52+4.330127018925i)x=52+4.330127018925i}

So...we have no real solutions to this "rational" function

2 Jun 2014