Crack an Intriguing Radical Rational Equation | Can You?

Crack an Intriguing Radical Rational Equation | Can You?

🔍 Ready for a mathematical challenge? Join me as we dive deep into the world of radical rational equations! 🧠 In this intriguing exploration, we'll crack open a complex equation and unravel its secrets step by step. Can you keep up with the twists and turns? 🤔 Sharpen your math skills and join the journey to solve this puzzle together. 💡 Let's crack the code of an intriguing radical rational equation – can you rise to the challenge? 🎓✨ #mathematics #Equations #PuzzleSolving #MathChallenge #RadicalEquation #ProblemSolving #CrackTheCode #CanYouSolveIt

Topics covered:
Algebra Challenge
Rational Equation
Radical equation
How to solve Rational Equations?
How to solve Radical Equations?
Math Olympiad
Algebra
Math Tricks
Algebraic identities
Algebraic manipulations
Substitution
Real Solutions
Quadratic formula
Quadratic equations
Math Tutorial
Algebraic Challenging Equations
Math Olympiad Preparation

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[NadiehFan]

The key observation here is that

(x + 10) + (√x − 10) = (x + √x)

Let x + 10 = a, √x − 10 = b, then x + √x = a + b. Note that x + √x is the denominator of both fractions at the left hand side, so we must have x + √x ≠ 0, that is, a + b ≠ 0. Multiplying both sides of the equation by (x + √x)³ we have

(x + 10)³ + (√x − 10)³ = (x + √x)³

which is

a³ + b³ = (a + b)³

so

a³ + b³ = a³ + b³ + 3ab(a + b)

and therefore

3ab(a + b) = 0

but a + b ≠ 0 so we have

a = 0 ⋁ b = 0

that is

x + 10 = 0 ⋁ √x − 10 = 0

x = −10 ⋁ x = 100

The solution x = −10 should not be rejected if only real solutions are sought since it was not stated at the start that √x should also be real, because, after all, −10 itself _is_ a real number.

[ericerpelding]

Using Desmos to plot the left hand side produces a graph that is exttremely flat.

[walterwen]

Crack an Intriguing Radical Rational Equation:
[(x + 10)/(x + √x)]³ + [(√x – 10)/(x + √x)]³ = 1, x ꞓR; x = ?
x ꞓR; x > 0, x + √x > 0, x + 10 > 0
Let: y = x + 10, z = √x – 10; y > 0, y + z = x + √x > 0
[(x + 10)/(x + √x)]³ + [(√x – 10)/(x + √x)]³ = [1/(y + z)³](y³ + z³) = 1
[1/(y + z)³][(y + z)(y² – yz + z²)] = 1, y² – yz + z² = (y + z)² = y² + 2yz + z², 3yz = 0
yz = 0, y > 0; z = 0 = √x – 10, √x = 10; x = 10² = 100
Answer check:
[(x + 10)/(x + √x)]³ + [(√x – 10)/(x + √x)]³ = [(100 + 10)/(100 + 10)]³ + 0 = 1; Confirmed
Final answer:
x = 100

[bobbyheffley]

Since the square root of x is in the denominator, let x be positive.

[wes]

x=a^2=–10 is a real root of the original equation and should not be rejected.