Can You Crack This Amazing Radical Equation? | Algebra

Can You Crack This Amazing Radical Equation? | Algebra

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In this algebraic video, we explore an interesting radical equation. This problem is an excellent exercise for anyone preparing for Math Olympiad. Follow along as we break down the steps to solve this radical equation, and try to solve it yourself before we reveal the solution.

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📌 Topics Covered:

Algebra
Radical equation
Problem-solving strategies
Substitutions
Quadratic formula
Factorization
Quadratic equation
Algebraic identities
Real solutions
Complex number
Complex solutions

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[RashmiRay-cy]

The given equation can be rewritten as [(12x^2+1)/x]^1/3 + [(16x^2-1)/x]^1/3= 4 x^1/3. Thus, 4 = u+v, where u=(12+1/x^2)^1/3 and v = (16-1/x^2)^1/3 and u^3+v^3=28. Thus, uv=3 which implies (u, v)=(3, 1), (1, 3). If v=1, we get x=+/-1/(√15). If u=1, we get x=+/-i/(√11).

[Quest]

Eqn can turns to x[165x^4+4x^2-1]= 0 or
x (15x^2-1)(11x^2+1)= 0 or
x= 0; +-√1/15; +-i√1/11 ;
since x # 0 &-1/4 ;
for real x= +-(1/√15) solns.

[gregevgeni]

Obviously x ≠ 0 and x ≠ -1/4.
The given equation is equivalent to
³√((12x²+1)/(4x²+x)) + ³√((4x-1)/x) -
- ³√((64x/(4x+1)) = 0 (1).
Let t = ³√((12x²+1)/(4x²+x),
r = ³√((4x-1)/x) and s = ³√((64x/(4x+1) and the (1) rewrite as t +r + s = 0 and from the known identity
t³ + r³ + s³ =3 •t•r•s =>
(12x²+1)/(4x²+x) + (4x-1)/x +(-(64x/(4x+1))= 3³√{[(12x²+1)(4x-1)(-64x)]/
/[(4x²+x)(x)((4x+1)]} or after some algebra =>x = ± 1/√15 real solutions.

[walterwen]

Crack This Amazing Radical Equation:
³√[(12x² + 1)/(4x² + x)] + ³√[(4x – 1)/x] = ³√[64x/(4x + 1)]; x =?
4x² + x ≠ 0; ³√(12x² + 1) + ³√[(4x – 1)(4x + 1)] = ³√(64x²)
³√(12x² + 1) + ³√[(16x² – 1) = ³√(64x²), Let: y = 4x², 4x² + x ≠ 0; y ≠ 0
12x² = 3y, 16x² = 4y, 64x² = 16y; ³√(3y + 1) + ³√(4y – 1) = (³√16y)
[³√(3y + 1) + ³√(4y – 1)]³ = [(³√16y)]³
(3y + 1) + (4y – 1) + 3{³√[(3y + 1)(4y – 1)]}[³√(3y + 1) + ³√[(4y – 1)]
7y + 3[³√(12y² + y – 1)](³√16y) = 16y, ³√[(16y)(12y² + y – 1)] = 3y
(16y)(12y² + y – 1) = (3y)³ = 27y³, y(165y² + 16y – 16) = 0, y ≠ 0; 165y² + 16y – 16 = 0
165y² + 16y – 16 = (15y – 4)(11y + 4) = 0, 15y – 4 = 0 or 11y + 4 = 0
15y = 4, y = 4/15 = 4x², x² = 1/15, x = ± 1/√15 or y = – 4/11 = 4x², x = ± i/√11
Answer check:
³√(12x² + 1) + ³√[(16x² – 1) = ³√(64x²)
x = ± 1/√15, x² = 1/15
³√(12/15 + 1) + ³√[(16/15 – 1) = 4/(³√15) = ³√(64x²); Confirmed
x = ± i/√11, x² = i²/11 = – 1/11
³√(– 12/11 + 1) + ³√[(– 16/11 – 1) = – 4/(³√11) = ³√(64x²); Confirmed
Final answer:
x = 1/√15; x = – 1/√15; Two imaginary value roots, x = i/√11 or x = – i/√11