A Nice Radical Algebra Equation | Can You Solve? | Math Olympiad

A Nice Radical Algebra Equation | Can You Solve? | Math Olympiad

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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
Pascal Triangle
Algebraic identities
Problem-solving strategies
Quadratic equations
Math Olympiad
Substitution
Real solutions

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[ericerpelding]

Note that if a and b are two expressions, then a solution for the equation a^(1/4) + b^(1/4) = (a+b)^(1/4) can be found by setting either a or b equal to 0.

[RashmiRay-cy]

Let a = 33-7x and b = 31-9x. a, b are greater than or equal to zero. The given equation then becomes a^1/4 + b^1/4 = (a+b)^1/4 > a=b = a+b + 2 (ab)^1/4 [ 2 (a^1/2 + b^1/2) + 3 (ab)^1/4] > 2 (ab)^1/4 [ 2 (a^1/2 + b^1/2) + 3 (ab)^1/4] = 0. Now, [ 2 (a^1/2 + b^1/2) + 3 (ab)^1/4] is not zero. Thus, ab=0 > a=0, i.e., x=33/7 or b=0, i.e., x=31/9. So, x = 33/7, 31/9.

[RyanLewis-Johnson-wqxs]

Surd[(33-7x), 4]+Surd[(31-9x), 4]=2Surd[(4-x), 4] x=31/9=3 4/9=3.4 recurring It’s in my head.

[ericerpelding]

Note that the right side requires x<4.
Let A=8(4-x) and B=x+1 and also C=B/A, then
(A-B)^(1/4) +(A+B)^(1/4) = (2A)^(1/4), or
(1-C)^(1/4) + (1+C)^(1/4) = 2^(1/4)
The accepted solutions for C are -1, +1.
If C=-1 we have B/A = -1 which will give x = 33/7 > 4 and so is rejected.
If C=1 we have B/A = 1 which will give the solution x = 31/9 = 3.4444.

[gregevgeni]

Obviously 33 - 7x ≥ 0 and 31 - 9x ≥ 0
We observe that
2( 4 - x)^1/4 = (2^4(4 - x))^1/4 =
= ( 16(4 - x))^1/4 = (64 - 16 x)^1/4 =
=[(33 - 7x) +(31 - 9x)]^1/4 .
That's the key of solution .
The original equation is equivalent with the
(33 - 7x)^1/4 + (31 - 9x)^1/4 =
= ((33 - 7x) + ( 31 - 9x))^1/4 ... (*)
Set 33 - 7x = t and 3 - 9x = s => (*) :
t^1/4 + s^1/4 = (t + s)^1/4 etc etc.
Roots, x = 33/7, x = 31/9 .

[АндрейПергаев-зн]

Не верно что 2а^2+2b^2+3ab=0 имеет только комплесные корни
а=b=0, выражение справедливо и корни не комплексные