An Amazing Radical Math Equation | Algebra Challenge

An Amazing Radical Math Equation | Algebra Challenge

Welcome to another exciting Math Olympiad challenge! In this video, we solve a fascinating radical equation. Join us as we delve into this intriguing algebraic challenge and explore the elegant solutions. Whether you're a math enthusiast, a student preparing for competitions, or simply love solving problems, this video is for you. Let's solve this together and enhance our problem-solving skills!

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🔍 In this video:

Detailed walkthrough of a challenging algebra problem from the Math Olympiad.
Tips and tricks for solving complex algebraic equations.
Encouragement to enhance your problem-solving skills and mathematical thinking.

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[RashmiRay-cy]

√(9-x)=x^2-10x+21 > x^4-20x^3+142x^2-419x+432=0. Note that 432=(27)(16). Let us On comparing, a=-11 and b=-9. So either x^2-11x+27=0 > x = 1/2[11+/-√(13)] or x^2-9x+16=0 > x = 1/2[9 +/-√(17)]. Note that 9-x is positive. So, x = 1/2[11+√(13)], 1/2[9 + √(17)] are ruled out. It turns out that x=1/2[9-√(17)].

[souzasilva]

O que levou a calcular a média aritmética entre 3 e 7? como saber que isso se faz necessário? (What led to calculating the arithmetic mean between 3 and 7? How do you know that this is necessary?)

[gregevgeni]

9 - x > 0 and 3 - x ≠ 0
x < 9 and x ≠ 3 . (*)
The given equation equivalent to √(9-x) = (7-x)(3-x) ≥0 (**)
(√(9-x))^2 = (7-x)^2 • (3-x)^2 =>
9-x = (7-x)^2 • (3-x)^2 =>
5-x +4 = (5-x +2)^2 • (5-x -2)^2
w+4 = (w+2)^2 • (w-2)^2 ( 5-x =w)
w+4 = [(w+2)(w-2)]^2
w+4 = w^4 -8 w^2 + 16
w^4 - 8 w^2 - w +12 = 0
(w^2 -w-4)(w^2 +w -3) = 0
w =(1±√17)/2, w =(-1±√13)/2.
But 5-x = (1±√17)/2 => x = 5-(1±√17)/2 and 5-x = (-1±√13)/2 =>
x = 5-(-1±√13)/2 (#)
From (*) (3-x)(7-x)≥0 <=> x≤3 or x≥7 (***)
From (*) and (***) x<3 or 7≤x≤9 .
The roots (#) are accepted only if
roots <3 or 7≤ roots ≤ 9 . ..
Observation.
w^4 - 8 w^2 -4 w + 12 =
(w2 + a w + b)(w^2 + c w +d) =>
a+c = 0
b+d+ac =-8
ad + bc=-1
bd=12 . . .=>(a, b, c, d) =(-1, -4, 1, -3).

[RealQinnMalloryu]

(9 ➖ x)^2/3 ➖ x)^2=(81 ➖ x^2)/9 ➖ x^2)={x^0+x^0 ➖ }/{x^0+x^0 ➖ }=x^1/x^1=x^1 (x ➖ 1x+1).(x ➖ 9x+9)/x ➖ 3x+3) (x ➖ 3x +3)(x ➖ 3x +3)/(x ➖ 3x+3) (x ➖ 1x+1)(x ➖ 1x+1)/(x ➖ 1x+1) (x ➖ 1x+1). (7 ➖ x)^2=(49 ➖ x^2)=(x^0+x^0 ➖ )=x^1 (x ➖ 1x+1).