A tricky problem with a 'divine' answer!

Can you solve this equation? Thanks to Hrigved for the suggestion!

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[NestorAbad]

Those equations are usually quite tricky to solve, thanks for sharing!
My approach was the following: instead of making substitutions at the beginning, first I manipulated the equation,

√(x + 1/x) + √(1 - 1/x) = x
√((x²-1)/x) + √((x-1)/x) = x
√(x²-1) + √(x-1) = x√x

Now square both sides:

x²-1 + x-1 + 2√((x²-1)(x-1)) = x³

Expand the product inside the root and arrange terms:

2√(x³-x²-x+1) = x³-x²-x+2

At this point, the substitution is more clear: let's call y=x³-x²-x+1. The equation then transforms to

2√y = y+1

Square both sides and arrange terms:

4y = y²+2y+1
0 = y²-2y+1
0 = (y-1)²
y = 1

Back to the substitution:

1 = x³-x²-x+1
0 = x³-x²-x
0 = x(x²-x-1)

And since x=0 is not a solution of the original equation, we are left with x²-x-1 = 0. The ending is then like in the video.

[DallasMay]

It seems like someone started with the golden ratio and then worked backwards to find the most absolute complicated looking equation they could come up with.

[КузебайГерд-ые]

"You should be able to solve it"

Yeaa... and you should be able to perform the 500 kg deadlift.

[peterkwan]

I solved by:

Multiplying both sides by sqrt(x),
sqrt(x^2 - 1) + sqrt(x - 1) = x sqrt(x)
Moving sqrt(x-1) to the right side and taking square on both sides,
x^2 - 1 = x^3 + x - 1 + 2 x sqrt(x^2 - x)
Subtracting 1 from both sides and dividing both sides by x (since x is not equal to 0),
x^2 - x + 1 - 2 sqrt(x^2 - x) = 0
The left side is a perfect square, so we get :
x^2 - x - 1 = 0
Solving for x, we get:
x = (1 + sqrt(5))/2 or x = (1-sqrt(5))/2

However, x = (1-sqrt(5))/2 does not fit in the original equation since x will be negative but the two square roots are both positive. So x = (1+sqrt(5))/2 is the only solution.

[math]

I did not even know where to begin. This was complicated. Appreciate and understand your logic. Well done.

[PExHzLRuuiFMgXUv]

There is also an interesting geometric perspective to the problem.

You can construct two right-angled triangles with sides (a, 1/√x, √x) and (b, 1/√x, 1) where a and b are defined as in the video. Now construct a larger triangle by joining these two, so that they share the side with length 1/√x.

Since we know that a+b=x, this larger triangle has sides (√x, 1, x). You can prove that this triangle is right-angled (exercise for the reader). Then, by Pythagoras theorem, √(1+x) = x, which gives the correct solution.

It is interesting that this triangle (aside from scaling) is the only one with the property that the ratio of the hypothenuse to the largest leg is equal to the ratio of the largest to the smallest leg, and that ratio is √phi.

[MSJ_]

Saw "divine" in the title and knew straightaway the answer would be the golden ratio 😛

[nickmeale]

Presh: "It's quite a divine answer!"
Me: Quite!"
*way over my head*

[marco_gallone]

When I was doing it on my own I ended up trying to factor a sextic function. Well that’s 6 years of engineering down the crapper!

[frentz]

You can also do this problem directly, with no particular "heroics, " simply squaring both sides of the equation to eliminate square roots, simplifying to isolate the remaining square root, and then squaring both sides again. And it has a couple neat moments! #1 I multiplied on both sides by sqrt(x), just to be lazy. (This will create the false solution x = 0, so we'll keep that in mind for later.) Now it says

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

You can think of it as sqrt(A) + sqrt(B) = x^(3/2). #2 Now square both sides (caution; may create false solutions). We get

A^2 + B^2 + 2 sqrt(AB) = x^3,

so x^2 + x - 2 + 2 sqrt((x^2 - 1)(x - 1)) = x^3. #3 Naturally the next step, isolate the " sqrt(AB) " term and square both sides again. But now look! A wonderful thing happens: let's simplify before squaring both sides the last time. We get

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

Almost the exact same cubic!! Set alpha = x^3 - x^2 - x + 1 (this is super neat, so go ahead and give yourself a greek letter :). .. squaring both sides, from

2 * sqrt(alpha) = alpha + 1,

you get 4 alpha = alpha^2 + 2 alpha + 1, which (yep!) simplifies to 0 = alpha^2 - 2 alpha + 1, or 0 = (alpha - 1)^2, so we must have alpha = 1. Plus (bonus!!) check it out : now #4 setting

alpha = x^3 - x^2 - x + 1 = 1,

we get a SOLVABLE cubic! So x^3 - x^2 - x = 0, or x (x^2 - x - 1). Very cool. Plus #5 we know that x = 0 cannot be a solution, because the original equation contains expressions "1 / x". So that leaves x^2 - x - 1 = 0, or x = (1 +- sqrt(5)) / 2. Finally in the original equation if you plug in, you can see only

x = (1 + sqrt(5)) / 2

works.

[harrymatabal]

Presh bhaijee you are too brilliant..you go straight to the point without unnecessary working. God bless you

[gamingmusicandjokesandabit]

The golden ratio really is golden to solving the problem 🙂

[amdc]

instead of implication you may (at each step) work ny equivalence : that means a≥0 and b≥0 and when you square an equation you add a sign condition ... then in the end (for sure you can still check if it you works fine to be sure but) you don't need to check, you have already eliminated the negative solution because of the conditions at each step... {TY for the video !}

[usptact]

“divine” was a huge hint that golden ratio will be involved. I guessed it :)

[purrlckhlmes]

Bruuuuh my math knowledge and solving is like hell and I'm in 10 standard. I hate math related things but dude!! When i saw your video in my recommendation it looked interesting and the looks exactly matched the content. You are the first one in my life to make me see maths like a entertaining subject. Keep it up man the videos are awesome!!

[karamsalah]

Magically, you make everything just simple!
Great explanation!

[Patient]

one of my all time favorite "Divine" moments in the art of mathematics is multiplication by Zero.
I enjoy watching Presh's manipulation of the equation... my eyes get tired quickly watching the numbers flying across the equals sign 🤣😀

[BiscuitZombies]

Hmm, intriguing. I solved it in the following manner:
Rearrange to give:
√(x-1/x) = x - √(1-1/x), then square both sides and cancel common terms,
x = x^2-2x√(1-1/x) +1

Then do the intriguing substitution, u = √(1-1/x), and we see that x = 1/(1-u^2) by rearranging, sub this in:

1/(1-u^2) = 1/(1-u^2)^2 - 2u/(1-u^2) + 1, multiply through by (1-u^2)^2 and get rid of common terms, put everything to one side,
u^4+2u^3-u^2-2u+1 = 0, you can use any method here, but by symmetry we can postulate that this is a perfect square;
(u^2+u-1)^2 = 0 ==> u^2+u-1 = 0 ==> u = (-1 +/- √5)/2

Once subbing these to find x, we have that x = (1-√5)/2 and x = (1+√5)/2. We can sub these into the original equation to reject anything extraneous, and we have that x = (1+√5)/2 = phi.

[IntrovertBoyz]

Just another solution :
Take y = 1/x
Then we have -
√(x-y) +√(1-y) = x
√(x-y) = x - √(1-y)
Squaring both sides
x - y = x² + (1-y) -2x√(1-y)
Cancelling y from both side and rearranging the terms -
x² - x + 1 = 2x√(1-y)
Dividing both sides by x -
x - 1 + y = 2√(1-y) [•.• y = 1/x]
Again squaring both sides -
x² + 1 + y² -2x -2y + 2 = 4(1-y). [•.•xy =1]
Simplifying the terms -
x² + 1 + y² -2x + 2y - 2 = 0
=> ( -x + 1 + y )² = 0 [•.• 2 = 2xy]
=> -x + 1 + y = 0
Replacing y by 1/x -
-x + 1 + 1/x = 0
=> -x² + x +1 = 0
=> x² - x - 1 = 0
Rest follows.
I think this is straight forward / conventional solution. Taking y = 1/x was only for simplification.

[schrodingerbracat]

Note: x >_ 1 for both square roots to be valid.
rewrite original equation as
(1-1/x)^(1/2) = x - (x-1/x)^(1/2)
Squaring leads to
x² - x + 1 = 2sqrt(x² - x)
Letting u = x² - x leads to
(u-1)² = 0
so u=1 and we have
x² - x - 1 = 0
solving, and remembering that x>_1 gives
x = [1+sqrt(5)] / 2