Math Olympiad | A Nice Algebra Problem | A Nice Radical Problem

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Find the value of x?
How to solve √x-1/x +√1-1/x =x

In this video, we'll show you How to Solve Math Olympiad Question A Nice Radical Equation √x-1/x +√1-1/x =x
in a clear , fast and easy way. Whether you are a student learning basics or a professtional looking to improve your skills, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems.
#matholympiad #mathtricks #maths

[luminiferous]

I solved this problem before watching the video or reading any comments.

My solution for x, where sqrt(x-1/x) + sqrt(1-1/x) = x is as follows:

x^2 = x - 1/x + 1 - 1/x + 2sqrt(x-1/x)sqrt(1-1/x)

x^2 = x + 1 - 2/x + 2sqrt(x-1/x-1+1/x^2)

x^3 = x^2 + x - 2 + 2sqrt(x^3-x^2-x+1))

x^3 - x^2 - x + 1 = 2sqrt(x^3-x^2-x+1))-1

Let u^2 = x^3 - x^2 - x + 1

Then u^2 = 2u - 1 ==> u^2 - 2u + 1 = 0

Using the quadratic formula:

u = [2 +/- sqrt((-2)^2 - 4)]/2 = [2 +/- sqrt(4 - 4)]/2 = [2 +/- 0]/2 = 2/2 = 1

Therefore, x^3 - x^2 - x + 1 = u^2 = 1

Subtracting 1 from both sides yields: x^3 - x^2 - x = 0

Dividing both sides by x (x is not 0 since x = 0 is not a solution shown by inspection of the original equation) yields: x^2 - x - 1 = 0

Using the quadratic formula:

x = [1 +/- sqrt(1 + 4)]/2 = [1 +/- sqrt(5)]/2 (Note: [1 + sqrt(5)]/2 is the Golden Ratio)

x = [1 + sqrt(5)]/2 = and x = [1 - sqrt(5)]/2 = are possible solutions

Check by inspection:

sqrt(x-1/x) + sqrt(1-1/x) = x

plug in x =

+ =

1 + =

1 + =

= is True, so x = is a solution


plug in x =

+ + sqrt(1 + =

1 + =

= is False, so x = is not a solution


Therefore, the solution is x = [1 + sqrt(5)]/2 = = the Golden Ratio.

[arampak]

Well, on one hand you solve a complex multi step problem that only experienced people can tackle, on the other hand you explain each step as if for a five-grader, and without actually describing why you do that way, what made you think along these lines and come to the solution. Otherwise a nice solution. I have solved it by a chain of transitions without introducing new variables, but yours is more elegant for sure.

[makmathematician]

Sir from which app do you make Thumbnails...?