Solve Radical Equation Like a Pro! | An Algebra Challenge

Solve Radical Equation Like a Pro! | An Algebra Challenge

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In this algebraic video, we explore an exciting radical equation that’s sure to challenge our 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
Problem-solving strategies
Factorization
Substitutions
Quadratic equations
Quadratic formula
Binomial Expansion
Synthetic division
Extraneous solutions
Rational root theorem
Real solutions
Verification

Additional Resources:

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

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[gregevgeni]

Another approach.
p = ⁷√(27+x) and q = ⁷√(100-x) .
For the real p and q the identity holds
etc etc .

[ShriH-do]

Let a^7=27+x; b^7=100-x; => a+b=2-1; or -1+2; & a^7+b^7=128-1 or -1+128 = 2^7-1^7 or -1^7+2^7; =>27+x=128 or -1 => x =101 or -28; verified ✓

[guyhoghton]

Let _uₙ = aⁿ + bⁿ_
∴ _u₀ = 2, u₁ = 1, u₇ = 127_
Use: *_uₘ₊ₙ = uₘuₙ - uₘ₋ₙ(ab)ⁿ_*
_u₂ = u₁₊₁ = u₁² - u₀(ab)¹ = 1 - 2ab_
_u₄ = u₂₊₂ = u₂² - u₀(ab)² = (1 - 2ab)² - 2(ab)² = 1 - 4ab + 2(ab)²_
_u₃ = u₂₊₁ = u₂u₁ - u₁(ab)¹ = 1 - 2ab - ab = 1 - 3ab_
_127 = u₇ = u₄₊₃ = u₄u₃ - u₁(ab)³ = (1 - 4ab + 2(ab)²)(1 - 3ab) - (ab)³_
⇒ _7(ab)³ - 14(ab)² + 7ab + 126 = 0_
⇒ *_(ab)³ - 2(ab)² + ab + 18 = 0_*

[gregevgeni]

x = 101 or x = - 28
Put p = ⁷√(27+x) and q = ⁷√(100-x).
Then
p⁷+q⁷ = 127 (1) and p+q = 1 (2).
We have
p⁷+q⁷=(p³+q³)(p⁴+q⁴) - p³q³(p+q) (3)
Now
p³+q³=(p+q)³-3pq(p+q)= 1-3(pq) (4), due to (2).
Too p⁴+q⁴=(p²+q²)-2p²q² =
[(p+q)²-2pq]²-2p²q² = (1-2pq)²-2(pq)² = 1-4(pq)+2(pq)² (5), due to (2).
From (1), (2), (3), (4), (5) we have
p⁷+q⁷=(1-3(pq))•
•(1-4(pq)+2(pq)²)-(pq)³ =>
127 = -7(pq)³+14(pq)²-7(pq)+1 =>
m³ -2m²+m+18 =0 (6), m=pq.
(6) <=> (m+2)(m²-4m+9)=0 =>
m =-2 => pq =-2 (7).
From (2) and (7) => p, q roots of the equation t²-t-2=0 <=> (t-2)(t+1)=0 .
So (p, q)=(2, -1) or (p, q)=(-1, 2).
For p=2 => ⁷√(27+x)=2 => 27+x=2⁷=> x=101 .
For p=-1 => ⁷√(27+x)=-1 => 27+x=(-1)⁷ => x=-28.
So the solutions of the original equation are x=101 or x=-28 .