Solve Radical Equations FAST with These Proven Strategies!

Solve Radical Equations FAST with These Proven Strategies!

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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
Algebraic identities
Extraneous solutions
Real solutions
Verification

Additional Resources:

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

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[RyanLewis-Johnson-wqxs]

Surd[(126-xSqrt[x])^2, 3]=26-x x=1 x=25 x=13+Sqrt[102Sqrt[51]-579]

[RashmiRay-cy]

Let √x=a. a is positive. Then, x=1, 5 are roots. [ a^4+6a^3-8a^2-204a-170. This does not yield real values of x. So, x = 1, 25.

[gregevgeni]

Obviously x ≥ 0.
Initial equation <=>
x³ -39x² +1014x -126x√x-850 = 0 (1)
Let √x = u => x = u², x² = u⁴,
x³ = u⁶, then the (1) rewrite
u⁶ -39u⁴ -126u³ +1014u² -850 =0 =>
= 0 => u =1 or u=5 or
u⁴ +6u³ -8u² -204u -170 = 0
(Irreducible polynomial).
For u=1 => √x =1 => x =1 accepted.
For u=5 =>√x =5 => x =25 accepted.

[Fjfurufjdfjd]

Πρεπει χ>0. οποτε εχω χ^(3/2)+ψ^(3/2)=126.και χ+ψ=26. Ανχ^(1/2)=α και ψ=β^(1/2) με α; β>0 εχω:α^3+β^3=126 α^2+β^2=26. ... (αβ)^3-39(αβ)^2+850=0. Παιρνω αβ=5 ή αβ=[34+(1836)^(1/2)]/2>0. Η δευτερη τιμη του αβ δεν δινει πραγματικη λυση. Αρα αβ=5 και α+β=6>0 α=1 ; 5. Τελικα χ=1^2=1 ; χ=5^2=25.

[alessandromarcovaldi-vw]

if you insert the equation into geogebra you get another solution 25.223980....