A Nice Radical Equation | An Algebra Challenge | You Should Try!

A Nice Radical Equation | An Algebra Challenge | You Should Try!

Welcome to infyGyan! In this video, we present an interesting and nice radical equation that will test our algebra skills and challenge our mathematical thinking. This problem is perfect for those preparing for Math Olympiad or anyone who loves tackling hard math problems.

Join us as we break down the steps to solve this intriguing equation, and see if you can find the solution on your own. Don't forget to leave your answers and approach in the comments below!

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[RashmiRay-cy]

Squaring and rearranging, the given equation becomes x^4-5x^3+6x^2-5x+ 1=0 > x^2+1/x^2 -5(x+1/x) +6 = 0 > (x+1/x)^2-5(x+1/x)+4=0 > x+1/x=1, 4. x+1/x=1 does not yield real values of x. If x+1/x=4 > x^2-4x+1=0 > x= 2+/-√3. Both are valid solutions. So, x= 2+/-√3.

[michaeldoerr]

This is *much* easier than the comments make it out to be. Maybe it is because I have been watching a good amount of infyGyan videos!!!

[gregevgeni]

X ≥ 0 and [14/(x -3)] - [9/(x -2)] ≥ 0
Solving the system we have
1/5 ≤ x < 2 and x > 3 (*) .
The given equation equivalent to
x^2 = [14/(x -3)] - [9/(x -2)] <=>
x^4 - 5 x^3 + 6 x^2 - 5 x + 1 = 0 .
Divine the last equation with x^2. ( x^2 ≠ 0 due to (*)) =>
( x^2 + 1/x^2) - 5(x + 1/x) + 6 =0 (**)
Put x + 1/x = t (#).
The (**) due to (#) written
t^2 - 5 t+ 4 =0 => t = 1 and t = 4.
From (**), x^2 - x + 1 = 0 and
x^2 - 4 x + 1 =0 .
The x^2 - x +1 =0 have two complex roots and the x^2 - 4 x +1 = 0. have real roots x = 2 ± √2 both of them satisfy the condition (*).

[RealQinnMalloryu]

(14)^2/(x ➖ 3)^2 ➖, (9)^2/(x ➖ 2)^2= 196/(x ^2 ➖ 9) ➖ 81/(x^2➖ 4)= 115/{x^0+x^0 ➖ }{x^0+x^0 ➖ }=115/{x^1+x^1}=115/x^2 =57.1 3^19.1 3^19^1.1 3^1^1.1^1 3^1 (x ➖ 3x+1).

[SidneiMV]

x = √[14/(x - 3) - 9/(x - 2)]
x = √(14x - 28 - 9x + 27)/√[(x - 3)(x - 2)]

x = √(5x - 1)/√[(x - 3)(x - 2)]

x√[(x - 3)(x - 2)] = √(5x - 1)
x²(x - 3)(x - 2) = 5x - 1
x²(x² - 5x + 6) = 5x - 1

I don't know how to continue