A Challenging Radical Equation | Give It A Try!

[@Quest3669]

Let a&b the 1st n 2nd term .
Solving gives a= 2; -1& b= -1; 2 real solns. i e
x= -65/44; -17/11 solns

[@RashmiRay-c1y]

Let a=[(20x+31)/2x+2]^1/5 and b=[(42x+62)/(2x+3)]^1/5. Then, a+b=1 and a^5+b^5=31. So, ab=-2,3. ab=3 does not yield real solutions. If ab=-2, we get a=2,-1. If a=2, x=-65/44. If a=-1, x=-17/11. So, x=-65/44, -17/11.

[@RyanLewis-Johnson-wq6xs]

Surd[(20x+31)/(2x+3),5]+Surd[(42x+62)/(2x+3),5]=1 x=-1.54recurring=-17/11 x=-0.01(147.72recurring)=-65/44 It’s in my head.

[@潘博宇-k4l]

X1 =-65/44, X2=-17/11.

[@satyanarayana560076]

looking for a siimilar problem under Japanese Olympiad maths training.

[@ashokdubey8415]

x= -65/44 & -17/11

[@Fjfurufjdfjd]

Πρεπει χ=/-3/2. Εχω:(20χ+31)/(2χ+3)=[10(2χ)+31]/[(2χ)+3] και (42χ+62)/(2χ+3)=[21(2χ)+62]/[(2χ)+3]. Θετω 2χ=α και εχω [10(α+3)+1]/(α+3)=10+1/(α+3) και [21(α+3)-1]/(α+3)=21-1/(α+3). Αν 10+1/(α+3)=β^5 και 21-1/(α+3)=γ^5 τοτε η εξισωση γινεται:β^5+γ^5=31 εχω το συστημα β+γ=1 β^5+γ^5=31.....βγ=3; -2. Αρα β+γ=1 ; βγ=3 ή β+γ=1 ; βγ=-2.
Τελικα εχω α=-65/22 ; α=-34/11 οποτε
χ=-65/44 ; χ=-17/11.