The BEST Way to Solve This Radical Equation | Olympiad Algebra Prep

The BEST Way to Solve This Radical Equation | Olympiad Algebra Prep

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In this algebraic video, we explore an intriguing radical equation that’s sure to challenge our algebraic skills. This problem is an excellent exercise for anyone preparing for Math Olympiad or simply looking to deepen their understanding of advanced algebra. Follow along as we break down the steps to solve this complex equation, and try to solve it yourself before we reveal the solution. Perfect for students and math enthusiasts alike!

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📌 Topics Covered:

Radical equations
Algebraic manipulation
Problem-solving strategies
Quadratic equations
Exponents
Factor
Substitution
Real solutions
Verification

Additional Resources:

#math #radicalequation #algebra #problemsolving #education #matholympiad #matholympiadpreparation #tutorial #quadraticequations

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[RashmiRay-cy]

Let a = (9-√x)^1/3 and b= x^1/6. Then, a+b=3 and a^3+b^3 = 9. If ab=t, (a+b)^3 = 27 = a^3+b^3 + 3t(a+b) > 27 = 9 + 9t > t=ab=2. a+b=3 and ab -2 > b = 1, 2 > x=1, 64. Both are valid solutions.

[Chacal]

X >= 0, y³ = 9-Vx, y = 3-⁶Vx => ⁶Vx =(3-y) => Vx = -y³+9y²-27y+27 = 9-y³ => y²-3y+2 = 0 =>
(y-2)(y-1) = 9 => [y = 2 => 8 = 9-Vx => x = 1] or [y = 1 => 1 = 9-Vx => x = 64]

[gregevgeni]

Let (9 - √x)^1/3 = 3 - x^1/6 = y ( x > 0)
Then y^3 = 9 - √x and 3 - x^1/6 = y (*) =>
√x = 9 - y^3 and x^1/6 = 3 - y =>
√x = ( 9 - y^3) and ((√x))^1/3 = 3 -y => (9 - y^3)^1/3 = 3 - y =>
9 - y^3 = (3 - y)^3 =>
y^2 - 3y + 2 = 0 =>
y = 1 or y = 2 .
From (*) x = 64 or x = 1 .
Both of them are accepted.

[RyanLewis-Johnson-wqxs]

Surd[(9-Sqrt[x]), 3]=3-Surd[x, 6] x=1 x=64

[Shobhamaths]

Let a^3=9-√x;
x=(9-a^3) ^2
6th root x=³√(9-a^3) ;
(3-a) ^3=9-a^3;
a^2-3a+2=0
a=1, 2;
x=64, 1;👍

[kassuskassus]

Let a=(x)^1/6 and solve for a the equation a^2-3a+2=0. We'll get a=1 that gives x=1 and a=2 that gives x=64.

[АндрейПергаев-зн]

Куча не нужных действий
Если уже ввели замену a и b, можно было и без этого
а^3+b^3=(a+b)*((a+b)^2-3ab=9
a+b=3
3*(9-3ab)=9
Отсюда ab=2, a+b=3

[walterwen]

Olympiad Algebra Prep: ³√(9 – √x) = 3 – ⁶√x, x ϵ R; x =?
Let: y = ³√(9 – √x); y = 3 – ³√(9 – y³), ³√(9 – y³) = 3 – y, [³√(9 – y³)]³ = (3 – y)³
9 – y³ = 3³ – 3(3²)y + 3(3)y² – y³, 9y² – 27y + 18 = 9(y² – 3y + 2) = 0, y² – 3y + 2 = 0
y² – 3y + 2 = (y – 1)(y – 2) = 0, y – 1 = 0; y = 1 or y – 2 = 0; y = 2
y = ³√(9 – √x) = 1, 9 – √x = 1, √x = 8; x = 8² = 64
y = ³√(9 – √x) = 2, 9 – √x = 2³ = 8, √x = 1; x = 1
Answer check:
x = 64: ³√(9 – √x) = ³√(9 – 8) = 1, 3 – ⁶√x = 3 – ⁶√64 = 3 – 2 = 1; Confirmed
x = 1: ³√(9 – 1) = ³√8 = 2, 3 – ⁶√1 = 3 – 1 = 2; Confirmed
Final answer:
x = 64 or x = 1

[SidneiMV]

x = u⁶

∛(9 - u³) = 3 - u

9 - u³ = (3 - u)³
u³ - 9 = (u - 3)³
u³ - 9 = u³ - 27 - 9u(u - 3)
9u² - 27u + 18 = 0
u² - 3u + 2 = 0
(u - 2)(u - 1) = 0

u = 2 => *x = 64*
u = 1 => *x = 1*

[RealQinnMalloryu]

( x^3 )^2➖ (9 )^2➖ x= {x^9 ➖ 81} ➖ x= 72 ➖ (x)^2= {72 ➖ x^2}="70 2^35 2^5^7 2^1^1 2^1 (x ➖ 2x+1).
(3)^2➖ {x*x*x*x*x*x}={9➖ x^6} =3 (x ➖ 3x+3).