Radical Algebra Hacks You Can't Afford To Miss

[@АлександрАнискевич-у1ш]

First root = a, second root = b. So we have system: a+b=3, a^3+b^2=9. b=3,1,6;
(5x+4)/(x+1)=9,1,36.
We have x=-3/4, -32/31, -5/4.

[@Quest3669]

Manipulating gives
(4x+5)/(x+1)= 0; -27; 8 gives
X= -5/4; -32/31; -3/4 solns.

[@RajeshKumar-wu7ox]

X=-3/4, -32/31,-5/4

[@潘博宇-k4l]

X=-(3/4), -(5/4), -(32/31).

[@Fjfurufjdfjd]

Ειναι χ=/-1. Θετω (4χ+5)/(χ+1)=α^3 και (5χ+4)/(χ+1)=β^2.
Εχω α+β=3 και α^3+β^2=9 (με προσθεση των δυο παραπανω κατα μελη)
Με επιλυση του συστηματος εχω:
β=1 , α=2 ή β=3 , α=0 ή β=6 α=-3.
Με αντικατασταση εχω:
χ=-3/4=-0.75 ή χ=-5/4=-1.25 ή χ=-32/31=-1.032...
Ομως πρεπει (5χ+4)(χ+1)>=0 και χ=/-1.
Δηλαδη χ=-4/5=-0.80.ολες οι ριζες ειναι δεκτες.
Αρα χ=-3/4 ;-5/4 ; -32/31.

[@gregevgeni1864]

x ≠ -1
Let t= ³√((4x+5)/(x+1)) =>
t³=(4x+5)/(x+1) ..(1)
r =√((5x+4)/(x+1)) => r²= (5x+4)/(x+1) (2) .. (5x+4)/(x+1)≥0 (5x+4)((x+1)≥0 ( x t+r=3 (4).
Solving the system of the equations (2), (3) => r=3-t, t³+(3-t)²=9 =>
t=-3, t=0, t=2.
Case one, t=-3
From (1) => (-3)³= (4x+5)/(x+1)
-27(x+1)= 4x+5 = 31x=-32 x=-32/31 accepted due to (*)
Case two, t=0
From (1) => 0=(4x+5)/(x+1) x=-5/4 accepted due to(*)
Case three, t=2
From (1) => 2³=(4x+5)/(x+1)
8(x+1)=4x+5 4x=-3 x=-3/4 accepted due to (*) ..