A Very Nice Math Olympiad Problem | Can You Solve for x? | Algebra | Radical equation

[@SpencersAcademy]

Notice that on checking, x=1 does not satisfy the equation. Therefore, there are only two solutions: x=0 and x=4

[@captainteach007]

You don't mention how you calculated the decomposition of 6x^2 and 9x - to achieve the factorisation ...

[@stevendebettencourt7651]

When you square to solve, you need to check your solutions with the original equation. The x=1 solution does not work, but x=0 and x=4 do.

[@SpencersAcademy]

Yeah. Absolutely ❤

[@dusanpokorny329]

Ano. Umocňování není ekvivalentní úprava rovnic a zde se provádí 2x. Ve výsledku se mohou objevit "falešné kořeny" zde dvojnásobně x=1. Zkouška u těchto iracionálních rovnic je povinná.

[@Sitthijack]

Thank you

[@SpencersAcademy]

You're welcome 😊

[@key_board_x]

x = √[3x + √(4x)] → where: x ≥ 0 ← we can see that: x = 0 is a solution
x² = 3x + √(4x)
x² - 3x = √(4x)
(x² - 3x)² = 4x
x⁴ - 6x³ + 9x² = 4x
x⁴ - 6x³ + 9x² - 4x = 0
x.(x³ - 6x² + 9x - 4) = 0
First case: x = 0
Second case: (x³ - 6x² + 9x - 4) = 0
x³ - 6x² + 9x - 4 = 0
x³ + 9x - 6x² - 4 = 0
(x³ + 9x) - (6x² + 4) = 0 ← we can see here that: (x = 1) is a solution, so we can factorize (x - 1)
x³ - 6x² + 9x - 4 = 0 → you factorize (x - 1)
(x - 1).(x² + βx + 4) = 0 → you expand
x³ + βx² + 4x - x² - βx - 4 = 0 → you group
x³ + x².(β - 1) + x.(4 - β) - 4 = 0 → you compare with: x³ - 6x² + 9x - 4 = 0
For x² → (β - 1) = - 6 → β = - 5
For x → (4 - β) = 9 → β = - 5
(x - 1).(x² + βx + 4) = 0 → where: β = - 5
(x - 1).(x² - 5x + 4) = 0 → then we solve for x the equation:
x² - 5x + 4 = 0
Δ = (- 5)² - (4 * 4) = 25 - 16 = 9
x = (5 ± 3)/2
x = 4 or x = 1
Summarize:
x = √[3x + √(4x)]
x.(x³ - 6x² + 9x - 4) = 0
x.(x - 1).(x² - 5x + 4) = 0
x.(x - 1).(x - 4).(x - 1) = 0
x.(x - 1)².(x - 4) = 0
Solution = { 0 ; 1 ; 4 }
x = √[3x + √(4x)] → when: x = 0
0 = 0 ← true
x = √[3x + √(4x)] → when: x = 1
1 = √[3 + √(4)]
1 = √[3 + 2]
1 = √5 ← false ← 1 rejected
x = √[3x + √(4x)] → when: x = 4
4 = √[12 + √(16)]
4 = √[12 + 4]
4 = √16
4 = 4 ← true
Solution = { 0 ; 4 }

[@SpencersAcademy]

Excellent delivery 💯 ❤

[@SGuerra]

A questão é das boas. Parabéns pela escolha. Brasil Novembro de 2024.

[@SpencersAcademy]

Thanks❤

[@davidshen5916]

Let Y=4X+Sqrt（4X）， Y-X=3X+Sqrt（4X）， X=Sqrt（Y-X），X^2=Y-X， （Y-4X）^2=4X， （X+Y-4X）（X-Y+4X）=Y-5X， （5X-Y）（Y-3X+1）=0， Y=5X，X^2=4X，X=0，X=4，Y=3X-1，X^2=2X-1， X=1

[@SpencersAcademy]

Excellent

[@msconsult3705]

Only 2 solutions: x=0 x=4

[@SpencersAcademy]

Absolutely

[@karimMalim-k6e]

Number 1 IS not solution

[@SpencersAcademy]

You're absolutely correct ❤