A Radical Equation | Math Olympiads

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[anastasissfyrides]

About a year ago i remember i came across this in a facebook group of relatively low level what so called math challenges and was stunned as it was significantly harder than anything id seen prior in the group, so naturally i became invested in it. After reaching it to the standard polynomial form step i had to factorise a quartic and thats when i gave up as i had no clue after some hours. So Wolfram gave me the numeric solutions and only then did i know how it was possible to be factorised. Havent seen your solution yet though

[andylee]

Always satisfying when terms fall out. :)

[goldfing]

First distribute the two roots evenly to both sides:

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

Then square the equation:

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

Add 1/x and 2*x*sqrt(1 - 1/x) and subtract x:

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

Square the equation once again:

4x^2 * (1 - 1/x) = x^4 + x^2 + 1 - 2x^3 + 2x^2 - 2x

4x^2 - 4x = x^4 - 2x^3 + 3x^2 - 2x + 1

We get the quartic equation

x^4 - 2x^3 - 1x^2 + 2x + 1 = 0

which is unfortunately not symmetric.
But we can use Bombelli's method:

x^4 - 2x^3 = 1x^2 - 2x - 1

Now (x^2 - x + z)^2
= x^4 + x^2 + z^2 - 2x^3 + 2zx^2 - 2zx
= x^4 - 2x^3 + (2z + 1)x^2 - (2z)x + z^2

So we add (2z + 1)x^2 - (2z)x + z^2 to our quartic.
The left side vecomes a perfect square then:

(x^2 - x + z)^2 = (2z + 2)x^2 - (2z + 2)x + (z^2 - 1)

(x^2 - x + z)^2 = 2(z + 1)x^2 - 2(z + 1)x + (z + 1)(z - 1)

It is easy to see that for z = -1,
the right side gets a perfect square, too, namely zero.
We end up with

(x^2 - x - 1)^2 = 0

So we have two double solutions, both originating from the quadratic

x^2 - x - 1 = 0

as in the video.

[homathgaming]

I remember this question, it's from Canadian Mathematical Olympiad 1998!

I would use the second method, but at 1:33, instead of squaring, I would multiply by x, thus getting a common term of x³ - x² - x + 1 (use 2 = 1 + 1 for the right hand side).

Assuming a = sqrt(x³ - x² - x + 1), the equation may be converted to a² - 2a + 1 = 0, thus a = 1. This implies x³ - x² - x = 0.

Since x =/ 0 as x is a denominator, we have x² - x - 1 = 0, thus x = ½(1±sqrt(5)).

[thegiganerd]

5:29 you could've put everything on one side and factorize to ( √(t)-1)² =0 it would've been easier

[GlorifiedTruth]

"Let's start with the second one." But that makes it the first one. 🙂

[alnitaka]

I did it by sqrt(x-1/x) = x - sqrt(1-1/x) and square both sides. 1/x cancels, giving x^2-x+1 = 2x*sqrt(1-1/x). Square both sides, x^4-2x^3+3x^2-2x+1 = 4x^2(1-1/x) = 4x^2-4x., and this yields x^4-2x^3-x^2+2x+1 = 0. Substitute x = y + 1/2, and this gets (after multiplying by 16) y^4-(5/2)y^2+25/16 = 0. This is (y^2-5/4)^2 = 0, yielding roots of sqrt(5)/2, sqrt(5)/2, -sqrt(5)/2, and -sqrt(5)/2. Going back to x gives x = (1+sqrt(5))/2 = phi (twice) and x = (1-sqrt(5))/2 = -1/phi, which we can throw out as this causes complex numbers.

[giuseppemalaguti]

Elevo al quadrato e moltiplico per x(quindi x diverso da 0)..risulta un'unica soluzione t=1...per cui (ko)...x=0 (ko)

[sergeyzelensky]

From equation x is equal the sum of two square roots and >=0. We have only one solution!

[moeberry]

The only solution is the golden ratio.

[aminimam]

The 2nd method is easier😊. No need to square both sides. Take X^2 as a common denominator you will end up with : 2*SQRT( X^3-X^2-X+1)/Sqrt(x^2) = X^2-X+(2/x )-1
Sqrt(X^2)=x Thus multiply both sides with x
2*SQRT( X^3-X^2-X+1)=X^3-X^2-X+2
Make t=X^3-X^2-X+1
2 SQRT (t)= t+1 squaring both sides 4t=t^2+2t+1 so t^2-2t-1=0 or (t-1)^2=0 Thus t=1
X^3-X^2-X+1 = 1
X^3-X^2-X=0 X(X^2-X-1)=0 X=0 is rejected
X^2-X-1=0 x=(1+Sqrt(5))/2 is the solution

[lawrencejelsma]

Good problem! I like your painful extraneous roots thoughts into this problem. 👍 To not spawn lazy student studies I won't tell how x as phi golden ratio symbol and -1/phi as the other x possible solution (extraneous?? Use a calculator to be sure folks!!! 😂🤣)👍👍