Don't Even Think of Squaring Both Sides!

[@Prypak]

x=-4 doesn't work, you'd get 5 = -3, which obviously isn't true. That is because according to your geometrical construction, it wouldn't make sense for x to be negative, that wouldn't create any kind of triangle if it was the case.

[@singularityfn]

Squaring created an extra root

[@Jan_Lei.]

for x=-4 i get: 5 = -3 and that seems to be not real 😉

[@Prypak]

@Jan_Lei. Well you have 12/x that gives you -3, to which you subtract -4, so you add 4 and it's √7, I don't really know how you get your 5 from there on

[@Prypak]

@@singularityfn exactly, but my point is that you can disregard -4 right away because the way the AI voice's put it is using geometry, and if we were to allow the sides to have a negative length, it'd be going the opposite way, and you wouldn't have any triangle

[@theosmit6366]

@Prypak 12/-4 = -3; -3 - (-4) = 1. Sqrt(1) = 1

[@SUICR7AF]

If x=-4, the right side is less than 0 while the left is more or equal 0

[@Zomsteve]

You're right but in algebraic terms, both values are acceptable He only used geometry to simplify the problem

[@JamesMcCullough-lu9gf]

@Zomsteve Geometry is somewhat based off of algebra. Anything you do in geometry will be acceptable in algebraic terms, and the root -4 is not acceptable because literally checking it with the original equation proves that it is extraneous. Therefore, only 3 is acceptable.

[@princianorvz]

No need for a solution, 3 is one correct answer.

[@Malekthebirb]

Great way to solve the equation without having to go through an annoying quartic! You probably do know this already but -4 is an extraneous solution it would be great if you included that in the video. Much love

[@brain_station_videos]

Thanks, you're right. I should've addressed that.

[@AnotherOne-s8v]

You need to check the roots of x. If x = 3: sqrt(12/3 - 3) + sqrt(12 - 3) = 12/3 sqrt(1) + sqrt(9) = 4 1 + 3 = 4 which is correct If x = -4: sqrt(-(12/4) + 4) + sqrt(12 + 4) = - (12/4) sqrt(1) + sqrt(16) = -3 1 + 4 = -3 which is incorrect

[@musicsubicandcebu1774]

One root can be found easily by noting that (12/x) - x must be greater than, or equal to, 0 Solving the inequality gives x = 1, 2 or 3. Checking each in turn makes x = 3 a winner.

[@fhffhff]

0,125(cos(15x)+cos(7x)+cos5x+cos3x)+0,125(cos(21x)+cosx)+0,125(cos(11x)+cos9x)

[@ShivanshGupta-ew7ev]

Wow at 0:52 bro assumed 12/x = H² which means H = square root(12/x) but he assumes P as square root[(12/x) - x] even though it should be P²

[@fhffhff]

0,125sin(21x)/21-0,125sin13x/13+c

[@CAPTAIN-1234.]

Even the Google saying unable to solve 😂

[@Garensonic]

Was it given in the question that x is real? If so then x = -4 is extraneous since the domain of the equation is that x > 0 and x^2

[@عمرحسينسيد]

You have forgot that the sum of two real sqare roo

[@alucardthespy5539]

You could also solve this with logic... 12 - x must be a square... x must be a divisor of 12... so 1, 2, 3, 4, 6, 12 (+ or -) 12 - 1 = 11, NO 12 - -1 = 13, NO 12 - 2 = 10, NO 12 - -2 = 14, NO 12 - 3 = 9, YES 12 - -3 = 15, NO 12 - 4 = 8, NO 12 - -4 = 16, YES 12 - 6 = 6, NO 12 - -6 = 18, NO 12 - 12 = 0, YES, But No, because 12/x - x would be negative 12 - -12 = 24, NO x = 3, x = -4 Double check... √(12/3 - 3) + √(12 - 3) ≟ 12/3 √(4 - 3) + √(9) ≟ 4 √(1) + √(9) ≟ 4 √(1) = ±1 √(9) = ±3 1 + 3 ≟ 4, Correct, ∴ x = 3 is a valid solution √(12/-4 - -4) + √(12 - -4) ≟ 12/-4 √(-3 + 4) + √(12 + 4) ≟ -3 √(1) + √(16) ≟ -3 √(1) = ±1 √(16) = ±4 1 + -4 = -3, Correct, x = -4 is a valid solution

[@musicsubicandcebu1774]

(12/x) - x, since under a radical sign must be greater than, or equal to, 0 . . . So x = 1, 2, or 3

[@ankurclimbs]

Your animation is best on youtube.

[@Radii_20]

That's...... amazing for such an equation is this method applicable for higher degree equations?

[@PramodShukla-wd9eh]

You are only who make me maths understand Thanks ❤❤❤❤❤❤❤❤❤❤❤❤🎉🎉🎉🎉

[@brain_station_videos]

You're welcome! Glad I could help.

[@PramodShukla-wd9eh]

@brain_station_videos thanks 🧡💝🧡😎🙏

[@রমজানআলী-ণ৫ঝ]

x^y=x^(y+1) x=? y=?

[@naturalsustainable6116]

Does your geometrical method shown return only real solutions which are not all possible solution,include complex solution?

[@JoãoSócratesAlmeida-n9t]

Drake

[@ronaksarda18s7]

Amazing method is this also applicable for other equations)?

[@LJR_NOOB69]

Great Method

[@prem198102]

You are Genius 😮

[@x.in_hype]

Good

[@PramodShukla-wd9eh]

Nice🎉❤❤❤❤❤❤😅

[@brain_station_videos]

Thanks 😊

[@PramodShukla-wd9eh]

@@brain_station_videos Welcome 🤗

[@CalculusIsFun1]

x = -4 is an invalid solution. the sum of two principal square roots will never be negative.

[@tunneloflight]

3

[@SuperCreepo1809]

First, Please Pin(I love Math)

[@shalinisharma3651]

first