A challenging equation for all precalculus students!

[@bprpmathbasics]

Behind the scene: kzbin.info/www/bejne/n6jaZ4t4Zd13mZY

[@DavidMadeira29]

I'd go with X=1 because I'm dumb like that. Namastè.

[@Thunderboys-l1v]

And also I'm dumb like that : ‘hello’

[@HammamAlsalem-lx5zo]

Cheers

[@MyOneFiftiethOfADollar]

Self-deprecating humor is as old as the hills. You don’t have a chance in this life if you really believe you are dumb.

[@AlanAsher-q8f]

Same

[@rewazza]

@@MyOneFiftiethOfADollar Don't take life so seriously

[@FaerieDragonZook]

Rather than squaring both sides, I like substituting y = sqrt(x). It also immediately reminds you to delete negative values of y.

[@MyOneFiftiethOfADollar]

@@FaerieDragonZooksince sqrt(x)>=0, set 1/(2-x) >=0 to determine solution domain. This avoids the crutch of introducing an unnecessary substitution.

[@nanamacapagal8342]

Set y = sqrt(x). 1/(2 - y^2) = y 1 = y(2 - y^2) 1 = 2y - y^3 y^3 - 2y + 1 = 0 (y - 1)(y^2 + y - 1) = 0 y = 1, -1/2 ± sqrt(5)/2 y = sqrt(x), so y >= 0. y = 1 -> x = 1 1/(2-1) = 1/1 = 1 = sqrt(1) y = -1/2 + sqrt(5)/2 5 > 1 sqrt(5) > 1 sqrt(5)/2 > 1/2 -1/2 + sqrt(5)/2 > 0 OK to use this value of y x = 3/2 - sqrt(5)/2 1/(2 - (3/2 - sqrt(5)/2)) = sqrt(3/2 - sqrt(5)/2) 2/(4 - 3 + sqrt(5)) = -1/2 + sqrt(5)/2 2/(1 + sqrt(5)) = -1/2 + sqrt(5)/2 2(sqrt(5) - 1)/(5 - 1) = -1/2 + sqrt(5)/2 sqrt(5)/2 - 1/2 = -1/2 + sqrt(5)/2 y = -1/2 - sqrt(5)/2 Obviously less than 0 so discard this solution

[@benpotito]

I did the same way

[@Insightfill]

As soon as you squared each side I started getting nervous. ALWAYS check your answers if you do that, as it can introduce extraneous roots.

[@itsphoenixingtime]

Alternatively.... Let sqrt(x) = u 1/2-u^2 = u 1 = u(2-u^2) 1 = 2u - u^3 u^3 - 2u + 1 = 0 u^3 - u^2 + u^2 - 2u + 1 = 0 u^2(u-1) + (u-1)^2 = 0 (u-1) (u^2 + u - 1) = 0 u = 1 or u = -1+sqrt(5)/2 or u = -1-sqrt(5)/2 Since u = sqrt(x), exclude the last solution as sqrt(x) need to be > 0 sqrt(x) = x , x = 1 -1+sqrt(5)/2 = x, x = 3 - sqrt(5)/2 These are the only 2 solutions and you don't even need to square both sides so the extraneous solution is removed from the start

[@thisisanotherplaceholder]

truly a golden answer. worth 79 points, so perfect like an aurora borealis

[@milos_radovanovic]

I was very interested in the question of why we don't have a third complex solution and decided to play with WA and plot log(abs(z^2t+z^t-1)) in the complex plane for different values of t and It is quite interesting to see the negative of the two solutions slowly disappearing "just around the corner" of 180 when t goes to < 1 or doubling for t>1

[@tastyfood2020]

1/(2-x)=√x 1²/(2-x)²=(√x)² x³-4x²+4x-1=0 x³-x²-3x²+3x+x-1=0 x²(x-1)-3x(x-1)+1(x-1) =(X²-3x+1)(X-1) And use quadratic equation then solve I might be wrong but this is the answer And also the answer have to be less than 2 if it isn't then it isn't the answer

[@sarimsalman2698]

How does one even develop an intuition to see that? I would have never seen that method to factor out the (x-1)

[@tastyfood2020]

@sarimsalman2698 well😅 It is a common method you should ask your teacher for better explanation ☺️

[@nikolakosanovic9931]

​@@sarimsalman2698this is one of the rare cases I saw it. I saw it a little differently. I know (x-1)³=x³-3x²+3x-1 x³-4x²+4x-1= x³-3x²+3x-1-x²+x= (x-1)³-x(x-1) . . .

[@cubing7276]

​@@sarimsalman2698from the factor theorem, equality holds when x=1, so (x-1) is a factor

[@SatyamSingh-j5o]

​@@sarimsalman2698 1st see all the basic properties as base....and most valuable.... Then just think looking at the eqn...like ah what terms could have been there so that my life with this ques would be easy... Seeing the cube and the const term as one... You would think (x-1)³...just make it that way..voilaa

[@Qermaq]

Hmm. Seems a reasonably simple algebra problem to me. If I square both sides, I have to be on the lookout for extraneous solutions. But doing so and cross-multiplying we get the cubic x^3 - 4x^2 + 4x - 1 = 0. The coefficients and the constant sum to 0, so (x - 1) is a factor. (x - 1)(x^2 - 3x + 1) = 0. This means x = 1 and/or x = (3 +/- sqrt(5))/2. Checking these three, we discard the positive version in the second case. This is a positive number, so its square root is also positive, but it's greater than 2 so the left side will be negative. But both x = 1 and x = (3 -sqrt(5))/2 work in the original equation.

[@bobross7473]

Bro it’s supposed to be simple lmao. It’s not an Olympiad problem or something

[@HaronLua69]

Them: [3 - sqrt(5)] / 2 Me: 2 - phi

[@echuidor]

Me: phi^2 - 2*phi +1

[@steveodonnell3030]

Nice I posted the same solution after you did. Foiled again.

[@mrswoopie1206]

1/2-x = sqrt(x) 1 = 2sqrt(x)-x (Multiplied 2-x on second side to get rid off fraction) Let's add a new value t, where sqrt(x) = t, t>0 1= 2t-t^2 t^2-2t+1=0 Using modernized Vieta theorem (cuz discriminant is too long to calculate) t=1 So x=1

[@cmorris7104]

Could be wrong but I believe x = (3 + sqrt(5))/2 is also a solution IF you think of the square root as always having a positive and negative result (e.g. sqrt(4) = 2 but also -2). But I guess it depends on context and the exact rules are a bit confusing to me

[@cmorris7104]

If you graph y = 1/(2 - x) and y^2 = x you’ll see what I mean

[@K42U]

I think he had a short video a year ago clarifying the confusion about the sqrt. √(x) will always have a positive result - the principal root. x^2 - A = 0 will always have two roots, as both pos. and neg. roots make the equation true. PS: I tried typing the equation in Wolfram Alpha and in my scientific a calculator app. Both gave the same roots. I also tried typing sqrt(x) = -4 in WA, it says "no solution".

[@acters124]

This is true because the equation on the right with sqrt of x should be in |sqrt(x)| if we want to invalidate the negative values. I believe there is a problem with current way people calculate square roots as they do have two answers. using absolute symbols is necessary because we need one answer. though because of how square roots work you are correct. however every calculator even graphing ones only display the positive values. I think we have a big problem with properly handling this as a math community. though I feel like the way we handle this as of right now is how you notice the way we handle it for the quadratic formula. we intentionally have to use the ± symbol to signify using both possible answers from the square root. which I find silly that we need to make this distinction for this specific case instead of other cases where you can have two answers.

[@stupidteous]

for sqrt to be a function it is defined to limit the range to x>=0, it only returns 0 or positive numbers. if not it wouldnt be a function. same with inverse trig functions.

[@K42U]

Yeah. Otherwise, it wouldN'T be a *function* at all. It'd be a RELATION instead. In fact, y^2 = x is a RELATION, not a function. The defining feature of a function is that every X value should map to only one Y value.

[@kath2oo4]

I just squared, multiplied, and got a cubic equation. then I did synthetic division to pull out the factor (x-1) and did the quadratic formula. then I made sure that the answers I got plugging back into the original equation fit the domain. ie: 1 works, (3-√5)/2 works, but (3+√5)/2 does not as when plugging into 1/(2-x) you get a negative number, and the sqrt function cannot give you a negative number as an output. also now looking at the video that's apparently what you did, lol, happy I did it well!

[@igorfiuza887]

I thought of subtracting both sides by 1, then on the left there would be (x - 1) and on the right √x - 1. But x - 1 = (√x - 1)(√x + 1), so you have √x - 1 on both sides and can say that x = 1 is a solution. When you divide by √x - 1 for x ≠ 1, you’ll have (√x + 1)/(2 - x) = 1 For x ≠ 2, say y = √x y + 1 = 2 - y² y² + y - 1 = 0 Then you have the other two solutions

[@ASChambers]

Another great video. It really demonstrates how some answers can just jump out at out and others might be a bit more sneaky. It’s always worth checking that a possible answer actually works ❤

[@Silvar55x]

I didn't square in the beginning (as in 0:43). Then later on it was more obvious that the negative result for √x was fake and needed to be discarded. x√x - 2√x + 1 = 0 (√x)³ - 2√x + 1 = 0 (a cubic in terms of √x) Observing that x = 1 √x = 1 is a root, did long division to factor out (√x - 1) and got (√x - 1) [(√x)² + √x - 1] = 0 √x - 1 = 0 √x = 1 x = 1 √x = (-1 ± √(1² - 4 · 1 · 1)) / (2·1) = (-1±√5)/2 Discard √x = (-1-√5)/2 < 0 Left with √x = (-1+√5)/2 x = (-1+√5)²/2² = (3 - √5)/2

[@tony_kara]

X³-4x²+4x-1 could be factored very easily you do this: (x-1)(x²+x+1)-4x(x-1)=0 Then you take (x-1) as a factor (x-1)(x²-3x+1)=0 so either x-1=0 or x²-3x+1=0 so x=1 or after quadratic formula x=(3+-✓5)/2

[@DonRedmond-jk6hj]

Could just leave as is and consider as cubic in sqr(x). It's a little to figure out who/re really the correct roots.

[@Ensign_Cthulhu]

I think it's worth explaining in detail why sqrt(x) is always positive whereas y=x^2 yields both positive and negative answers.

[@randerson4009]

If you add the negative square root of x to your graph you can nicely show the extraneous solution at the intersection.

[@jmich7]

I would start stating that x cannot be equal to 2. I love all your videos anyway, no matter what. Also, it is so interesting! Because we could think there is a hidden root! Which is not the case apparently since, originally, we do not have a third degree equation.

[@clarfonthey]

I like these challenging problems, since even though I'm far past precalc, I tend to forget that I can solve problems like this. Good job!

[@Mutualititve]

This is a good reminder to always do the conditions for each problems first.

[@milos_radovanovic]

Why is this tricky: 5:35? Because the squaring step is unnecessary and confuses the sign of the third (extraneous) solution. Substitute t=sqrt(x), solve as a cubic and select solution for t>0: t^3-2t+1=0 /(t-1): x1=1 -> t^2+t-1=0 /(t>0) -> t=(-1+sqrt(5))/2 -> x2=t^2=1/4+5/4-sqrt(5)/2 =(3-sqrt(5))/2

[@ProactiveYellow]

Given that the LHS involves division, we know that 2-x≠0, that is, x≠2. Given that the RHS is always non-negative, we know that 2-x≥0, that is, 2≥x. Combine these facts to get the restriction x½(3+2)=5/2>2, so we can reject tha larger solution, giving us solutions of x=(3-√5)/2, 1

[@magattbeats]

That’s why my teacher always tells me to know the condition of existence, where solution can be. In this example, the condition of existence is x≠2 and x>=0

[@itsphoenixingtime]

Alternatively.... Let sqrt(x) = u 1/2-u^2 = u 1 = u(2-u^2) 1 = 2u - u^3 u^3 - 2u + 1 = 0 u^3 - u^2 + u^2 - 2u + 1 = 0 u^2(u-1) + (u-1)^2 = 0 (u-1) (u^2 + u - 1) = 0 u = 1 or u = -1+sqrt(5)/2 or u = -1-sqrt(5)/2 Since u = sqrt(x), exclude the last solution as sqrt(x) need to be > 0 sqrt(x) = x , x = 1 -1+sqrt(5)/2 = x, x = 3 - sqrt(5)/2 These are the only 2 solutions and you don't even need to square both sides

[@mazenzidieh]

very nice! thanks

[@shannonmcdonald7584]

Fairly easy until the end. Question: in polynomial long division I get remainders allot. I never know what to do with the equation after that.

[@steveodonnell3030]

(3 - sqrt(5))/2 is equal to 2 - φ φ is everywhere!

[@janmesh2332]

Because of the way the question is framed x must belong to [0, 2) 1/(2-x) = sqrt(x) Square both sides 1/(2-x)^2 = mod(x) The first info was imp because otherwise it would be a pain to solve with mod here. Since x belongs to [0, 2) 1/(2-x)^2 = x Solving further x^3 - 4x^2 + 4x - 1 = 0 Solving this equation x = 1 and x = [3 +- sqrt(5)] / 2 But x = [3 + sqrt(5)] / 2 ~= 2.11 which is >2 So, the only solutions are x = 1 and x = [3 - sqrt(5)] / 2

[@TheBedLump_Sans]

Here's my attempt at it before watching the video: 1/(2-x) = sqrt(x) Multiply both sides by 2-x 1 = sqrt(x)*(2-x) 1 = sqrt(x)*2 - sqrt(x)*x x = sqrt(x)*sqrt(x) Sub in sqrt(x)*sqrt(x) for x x*sqrt(x) = (sqrt(x)*sqrt(x))*sqrt(x) = (sqrt(x))^3 So now we have 1 = 2*sqrt(x) - sqrt(x)^3 Let y = sqrt(x) 1 = 2y - y^3 y^3 - 2y + 1 = 0 By inspection y = 1 is a solution Therefore by factor theorem (y-1) is a factor Polynomial division y^3 - 2y + 1 = 0 ÷ y-1 = y^2 + y - 1 (can't show steps here) Quadratic formula y = (-1 +- sqrt(1-4*1*-1))/(2) y = (-1 +- sqrt(5))/2 But y >= 0 as y = sqrt(x), so -1-sqrt(5)/2 isn't a solution since it's negative If y = sqrt(x) then x = y^2 So sub in for y to find x values: When y = 1, x = (1)^2, x = 1 When y = (-1 + sqrt(5))/2: x = ((-1)^2 + 2(-1)(sqrt(5)) + (sqrt(5))^2)/(2^2) x = (1 - 2*sqrt(5) + 5)/4 x = (6 - 2*sqrt(5))/4 x = (3 - sqrt(5))/2

[@TheBedLump_Sans]

So it turns out I overcomplicated things with y = sqrt(x) in the middle instead of at the start😅tbh I don't even know why I didn't bother to think of doing that at the start...i mean at least i got the right answer

[@ZiRR0]

the denominator is phi if you plug-in (3 + sqrt(5)) / 2?

[@ZiRR0]

and the negative reciprocal if it's - right

[@mikseryasta6369]

when i have to write (√x)^ 2 as abs value of x, and when can i write that sqrt and power cancels?

[@saer6596]

As the square root is defined only for non negative numbers and its result is also non negative, (sqrt(x))^2 is just x as is understood that x must be non negative, of course there is nothing wrong in writing |x| but it would be redundant But sqrt(x^2) is |x| for the reason exposed above and because we want sqrt to be a well defined function Note that there is a big difference between the root square of a number p, and the solutions to the equation x^2-p=0 Hope it clarifies your question

[@richardslater677]

Question please. I was surprised to learn recently that the square root of, say, 4 when written inside the square root radical, only outputs the positive value, ie 2 and not -2. You use this fact in this video. My question is, if I asked someone, verbally, without writing anything down, “what is the square root of 4”, is the answer “2” or is “2 and -2”. I know that x^2=4 gives both values for x but what if asked the other way round? Thanks.

[@Engy_Wuck]

The square root of any *number* is positive. By definition. But for x²=4 you don't ask for the square root. You ask "which number(s), when squared, give the result 4". That's a completely different question - one that can be solved by a square root, but only up to the abolute value of x (essentially: the distance from 0), because both +2 and -2 satisfy that question. Therefore the "real" result of taking the square root on both sides of x²=4 would be |x| = 2. Since we don't like to have those "absolute value" sgns to calculate with further we then add the additional knowledge that in ℝ the only possible values for the distance from 0 are in the positive and negative direction, therefore |x|=±x, and here therefore x²=4 --> |x|=2 --> ±x=2 --> x=±2 Now we are lazy and only write x²=4 --> x=±2 But it's not the 4 that makes the ±, it's the x². Btw: That absolute value is the reason you have to check your answers after squaring - you remove the sign of that thing squared, therefore always get two possibles back even when only one existed before squaring.

[@richardslater677]

@ Most kind of you to take the trouble to explain it. Thank you.

[@NotGleSki]

How dare you disrespect my complex roots

[@lpi3]

Show me

[@CalculusIsFun1]

(1/(2-x))^2 = x 1/x = x^2 - 4x + 4 x^3 - 4x^2 + 4x - 1 = 0 Notice, x = 1 is a solution. (x^3 - 4x^2 + 4x - 1)/(x-1) = x^2 - 3x + 1 the roots of the quadratic are 3/2 +- root(5)/2 Of those two solutions, 3/2 + root(5)/2 is greater than 2 and will thus result in a negative in the denominator of the 1/(2-x) which is not possible in the real for a square root to have negative. This means the only two real solutions are x = 1 and x = (3/2) - (root(5)/2)

[@ronaldking1054]

Old school rational solution: 1/(2-x) = (x)^(1/2), 0

[@delent-team]

Very interesting

[@vicar86]

Always investigate the domain at the beginning.

[@Bxvxnce]

Famous last words, "I forgot to check for extraneous roots 😔"

[@jamesharmon4994]

I saw the earlier video 😊

[@ChavoMysterio]

[1/(2-x)]=√x Domain: 00 -x>-2 x

[@vanessakitty8867]

Nice one.

[@cyruschang1904]

1/(2 - x) = ✓x 1 = x(x^2 - 4x + 4) x^3 - 4x^2 + 4x - 1 = 0 (x - 1)(x^2 - 3x + 1) = 0 x = 1, (3 - ✓5)/2 x cannot be (3 + ✓5)/2

[@mikeoxsor6183]

hi i do not understand why the square root function cannot ouput a negative number in this case, i know about principle square root but we are solving for x here so why can't you use the other solution

[@gerryiles3925]

Because the square root "function" only outputs non-negative values by definition, because a "function" can only output a single value for any particular input value...

[@trueriver1950]

Sounds like you don't understand the definitions of a function and therefore of of the sqrt sign. By definition sqrt always provides the positive. If it was capable of providing more than one answer it would not be a function, by the definition of a function. That is why sqrt(x) is not the same as x^0.5. x^0.5 emits two answers, and therefore is not a function: mathematicians call it a relation. That is also why bprp made a point of emphasizing that each side of the original equation is a function, making sure students know the exact mathematical meanings of the technical words.

[@MyOneFiftiethOfADollar]

@@mikeoxsor6183 the reason is it would not be a function as it would be a “horizontal parabola” along positive x axis. Does not pass vertical line test for functions You would have to content yourself with statements like sqrt(9) = +/- 3

[@MyOneFiftiethOfADollar]

@@trueriver1950 speaking of meanings of technical words, please share the technical meaning of “emit”

[@joeschmo622]

How does The Cat figure into the video? I was looking for him the whole time.

[@nikolakosanovic9931]

Or you can check both solutions and see if they work

[@xytean]

i tried to resolve it without using any formula and i got this: 1/(2-x) = sqrt x (1/2-x)² = x 1²/(2-x)² = x 1/2² - 2*2*x + x² = x 1/4 - 4x + x² = x 1 = x * (4 - 4x + x²) 1 = 4x - 4x² + x³ 1/4 = x - 4x² + x³ sqrt((1/4)/-4)² = x - x + x³ sqrt(-4/16)² = x³ sqrt(-1/4)² = x³ -1/4 = x³ cbrt(-1/4) = x cbrt(-1) / cbrt 4 = x okk maybe im stupid... i've just noticed that this gives me a complex number 😭 UPDATE - 2 MINUTES HAVE BEEN PASSED AND IVE NOTICED AN ERROR, LET ME RETAKE THIS FROM BEFORE THE MISTAKE. 1/(2-x) = sqrt x (1/2-x)² = x 1²/(2-x)² = x 1/2² - 2*2*x + x² = x 1/4 - 4x + x² = x 1 = x * (4 - 4x + x²) 1 = 4x - 4x² + x³ 1/4 = x - 4x² + x³ sqrt((1/4)/-4)² = x - x + x³ sqrt(-4/16)² = x³ sqrt(-1/4)² = x³ (here i shouldnt cancel the sqrt with that power of two) sqrt(1/16) = x³ sqrt 1 / sqrt 16 = x³ cbrt 1/4 = x cbrt 1 / cbrt 4 = x (here im gonna rationalizate this denominator) 1 / cbrt 4 * cbrt 4²/cbrt 4² = x cbrt 4² /4 = x

[@m.h.6470]

Solution: First: x ≠ 2 Second: Substitute √x with y 1/(2 - y²) = y |*(2-y) 1 = y(2 - y²) → y = 1 is an obvious solution -y³ + 2y = 1 |*-1 +1 y³ + 0y² - 2y + 1 = 0 Polyn. division by (y - 1) y² + y - 1 = 0 y = (-1 ± √5)/2 → since y = √x, y can not be negative therefore only y = (-1 + √5)/2 is valid As such, we can calculate the x value as: √x = y |² x = y² x₁ = 1² = 1 x₂ = ((-1 + √5)/2)² ≅ 0.382

[@troubledouble106]

I just put x=1 into the equation in my head and got the answer.

[@NH_RSA__]

Solutions restricted to real numbers?

[@trueriver1950]

For precalculus students, that sounds right to me. Or maybe schools in your country introduce complex numbers before calculus???

[@NH_RSA__]

@trueriver1950 I don't quite remember which course introduced imaginary numbers. It may have been an EE course or precalc. But that was in '78 so not sure.

[@redrokee1082]

Very fun

[@BestCosmologist]

Call me crazy, but x=(1/(2-x))^2 is the most useful expression.

[@GianMichelAbreuDelgado]

I failed one root

[@samukaze5810]

After 15 minutes of solving, x=1 🤦

[@nightgame2365]

Since we have the root of x the answer it can be - or + so we have three answers Unless the equation is talking about the positive root of x

[@JayTemple]

Showing that (3 + sqrt(5))/2 didn't work was simple because the rational side was negative. The most tedious part of this was showing that (3 - sqrt(5))/2 is a valid solution, which you didn't do.

[@Warcraft_Traveler]

Tedious ? 3÷2 is 1.5 so if you substract sqrt of whatever positive real number from 3, there is no way you can end up with 2-(3-sqrt of whatever)/2 being negative.

[@Engy_Wuck]

for (3-√5)/2 to be less than 0 the top would have to be less than zero, but √55, so √9>√5 and therefore 3-√5 is bigger than zero, too. To be complete you'd also have to show that it's not equal to 2, because that would make the denominator zero, but... 😛

[@jamesharmon4994]

He did approximate sqrt(5) in the video. Using that info proves 2 - ((3-sqrt(5)) / 2 is positive

[@bobross7473]

I have a feeling when you end up with two square root answers like that one of them will be extraneous and the other will be valid, but I don’t know how to prove that this is always the case

[@JayTemple]

@@Warcraft_Traveler I wasn't satisfied with it not being negative. I actually showed that it satisfied the equation.

[@holyshit922]

x = 1 , but is it the only one solution

[@yassineomri8173]

No( 3-√5)/3 is also a solution