Amazing Radical Equation That You Can't Afford to Miss

[@RashmiRay-c1y]

The given equation is rewritten as (8+1/x^2)^1/5+ (25-1/x^2)^1/5=3. Let a=(8+1/x^2)^1/5 and b = (25-1/x^2)^1/5. Then, a+b=3 and a^5+b^5=33. Setting ab=t, we get t^2-94+14=0. So, t=7,2. But t=7 does not yield real solutions. For t=2, a=1,2. But a=1 does not give a real x. So, a=2 and x = +/- 1/(2√6).

[@Shobhamaths]

x=1/(2√6) Taking lcm, and divide both sides by (x) ^1/5 then (8+1/x^2) ^1/5+(25-1/x^2) ^1/5=3 Let a^5=8+1/x^2 b^5=25-1/x^2 a+b=3 a^5+b^5=33 a^4-6a^3+18a^2-27a+14=0 Solve (a, b) =(1, 2) , (2, 1) x=1/(2√6)👍

[@gregevgeni1864]

The given equation is equivalent to ⁵√(8x²+1) +⁵√(25x²-1)=3⁵√x² =>(:⁵√x²) ⁵√(8-1/x²) +⁵√(25-1/x²)=3 (1). Let t = ⁵√(8+1/x²) and r =⁵√(25-1/x²) => t + r = 3 (2) due to (1). Also t⁵ + r⁵ = 33 (3). Solving the system of (2), (3) => (t,r)=(1,2) or (2,1). So ⁵√(8+1/x²)= 2 => 8+1/x²=32 =>=> 1/x²=24 => x=±1/√24=±1/(2√6). . .

[@Quest3669]

Eqn can be {8+(1/x^2}}^1/5+{25-(1/x^2)}^1/5= 3 gives a^5+b^5= 33 & a+b= 3 gives a=2;1 & b=1;2 or 1/x^2= -7; 24 or real X= +-√1/24=+-1/2√6 soln.

[@Fjfurufjdfjd]

χ=+ - [(6)^(1/2)]/12.

[@Fjfurufjdfjd]

Θα ηθελα για την καινουρια χρονια να ευχηθω στους υπευθυνους του καναλιου και στους λατρεις της αλγεβρας, ηρεμια, καλοσυνη, αφθονια σε καλα αισθηματα, περισσοτερες στιγμες ευτυχιας, λιγοτερο αγχος και σε μενα περισσοτερες ελευθερες ωρες για να ασχολουμαι με τα αγαπημενα μου μαθηματικα, μηπως και καταφερω να γλυτωσω απο το αλτσχαιμερ.

[@RealQinnMalloryu4]

2^3 2^3x^1+1/2^3x^1+x+2^3 2^3 ➖ 1/x 1^1 1^1^1x+/1^1x+2^1 1^3 ➖/ 23/ (x ➖ 3x+2). 243x/(5x )^2➖ (1 )^2=243x/{25x^2 ➖ 1}=243x/,24x^2=1.3x^2 1.1x^2 1x^2 (x ➖ 2x+1). I wish you a happy new years eve and a happy new years day.