Solving a Radical Equation with Unlike Roots

[@maxbow-arrow5931]

Nice solution, but the one mentioned at the start also works. After cubing both parts we get 2-x = 1 - 3sqrt(x-1) + 3(x-1) - (x-1)sqrt(x-1) Then we isolate the square roots and get (x+2)sqrt(x-1)=4x-4 Notice that the RHS is just 4(x-1), so after squaring we get (x+2)^2*(x-1)=16(x-1)^2 So once we move everything to the left, we see that (x-1) can be factored out, yielding (x-1)(x^2+4x+4 - 16x+16) = 0 Solving the quadratic gives x=2 and x=10, and the first bracket gives x=1

[@SyberMath]

Wow! I thought it was going to be messier

[@shmuelzehavi4940]

Same way.

[@agnibeshbasu3089]

nice! bring more videos like this and olympiad problems.

[@SyberMath]

Thank you! I will bring you lots of interesting problems...😉

[@x714n0____]

Very nice! I never saw this kind of substitution... ✌🏻

[@SyberMath]

Thank you!

[@usernamehere94]

@@SyberMath Exactly! I was educated in the United States and I was never expose to such a substitution either. Thanks!

[@osmanfb1]

I just used u^3 - 2-x as a single substition. It worked out u + sqrt(2-u^3-1) = 1 sqrt(1-u^3)=1-u sqrt((1-u)(1+u+u^2)) = sqrt(1-u) sqrt(1-u) now u=1 is one solution. if u is not 1 then divide both sides by sqrt(1-u) sqrt(1+u+u^2) = sqrt(1-u) u(u+2)=0 u=0 and u=-2 are also solutions. In therms of x x=2 , x=10, and x=1 (corresponding to u=1, this is without introducing extra unknowns.

[@elkincampos3804]

Note that 2-x=1-(x-1). Then put u=√(x-1).It follows that (1-u^2)^(1/3)+u=1. u-1=(u^2-1)^(1/3). Therefore u^2-1=(u+1)*(u-1)=(u-1)^3. Thus 0=(u-1)^3-(u+1)*(u-1)=(u-1)*(u^2-2*u+1-u-1)=(u-1)*(u)*(u-3). Thus u=0,u=1 or u=3 that is x=u^2+1=1, 2 or 10.

[@jeromemalenfant6622]

???? Simpler: Let y equal the first radical, (as he does). Then x=2-y^3 and the second term is sqrt(1-y^3), so sqrt(1-y^3) = 1-y. Square both sides, the 1 cancels out, and you get y^3 + y^2 -2y = 0. One solution is y=0, or x=2. Dividing by y you get the quadratic, y^2 + y -2 =0, with solutions y = 1 and -2, corresponding to x = 1 and 10. So the solutions are x= 1, 2, 10.

[@SyberMath]

Nice!

[@nerusfuho674]

This is really awesome, I like what the different radicals work together 👍

[@SyberMath]

Glad you like it!

[@misterdubity3073]

I used u=2-x to get rid of the cube root. Then w=sqrt(1-u). Same results for x in the end. My solution worked, but more messy and not as elegant.

[@enalaxable]

Still I think substitution is a bit easier: u^3=2-x , gives u*(u^2+u-2) =0. and same results for u* in {-2,0,1}, thus x* in {10,2,1}. It's typical strategy to get rid the highest radical first.

[@SyberMath]

I agree!

[@ulrichkaiser3794]

At 2:25 it is easier to add eq 2 and eq 3 , resulting in y^3 + z^2 = 1 ... Then, I replaced z^2 by (1-y)^2 and so on, resulting in y (y+2) (y-1) = 0 From y = { 0 , -2 , 1 } one can derive x = { 2 , 10, 1 }

[@ulrichkaiser3794]

Thank you very much ! 🥰🌻🍀

[@AnthonySpinelli-fe4vn]

My method is as follows: move the cube root to the RHS and negate the inside of the radical (as the function is odd) and square both sides so as to simplify the LHS. And upon doing so, you can substitute x-2 for, say, a variable y (y = x - 2). With this equation (I’ll use rt to mean cube root) y=2rt(y)+rt(y^2). The next step isn’t real rigorous, but you can divide out the cube root of y (and keep more that y=0 was the lost solution) and solve a quadratic in terms of the cube root of y. This leaves the three answers as y = -1, 0 and 8, which give the x values of x = 1, 2, and 10.

[@SyberMath]

That's cool!

[@chhromms.8138]

easy to understand and great explanation. perfect

[@SyberMath]

Glad you think so!

[@mrphlip]

At the start, you say that just isolating the cube root and cubing both sides would get too messy, but it's not _that_ bad... You end up with: 2 - x = (1 - sqrt(x-1))³ 2 - x = 1 - 3sqrt(x-1) + 3(x-1) - (x-1)sqrt(x-1) We then want to isolate the square root: 2 - x = 3x - 2 - (x+2)sqrt(x-1) 4 - 4x = -(x+2)sqrt(x-1) 4(x-1)/(x+2) = sqrt(x-1) (noting we can safely divide by x+2 as x=-2 is not a solution) 16(x-1)²/(x+2)² = x - 1 x = 1 is clearly a solution, but we can cancel out the (x-1) to find other solutions 16(x-1)/(x+2)² = 1 16x - 16 = x² + 4x + 4 x² - 12x + 20 = 0 (x - 2)(x - 10) = 0 x = 2 or 10 Since we squared both sides we need to test the solutions in the original equation, but all three fit as shown in the video.

[@SyberMath]

Nice!

[@aashsyed1277]

@@SyberMath WOW. WOW

[@davidseed2939]

At about 2:00 you say write everything in terms of x. But that's not what you want and thats not what you do. We had everything in terms of x. What you effectively do is to eliminate x. Adding the last two equations gives y^3 +z^2 = 1 Then you substitute y=(1-z) . Now we have everything in terms of z which nicely cancels. If we had written the equation in terms of y. It would also work. y^3+ (1‐y)^2 =1 y^3 +(y^2 -2y+1) =1 y(y^2 +y -2)=0 y(y-1)(y+2)=0 solutions at y=0,1,-2 X =1,2 ,10

[@SyberMath]

You're right!

[@theafs4547]

Good solution

[@SyberMath]

Thank you, Mr. Einstein! 😁

[@indrjeetkumar1030]

Love your work 🔥

[@SyberMath]

Thanks 🔥

[@tharanathakula3588]

Excellent.

[@SyberMath]

Many thanks!

[@Deepak0Aggarwal]

Substitute is his favourite method

[@jonyp1320]

nice ! learned something

[@SyberMath]

Glad to hear that!

[@aliasgharheidaritabar9128]

Wonderful.you keep me busy with your videos.

[@SyberMath]

Glad you like them!

[@quajutsu9372]

Wherever I see a Math channel I just click on Subscribe, as long as I respect its owner like you

[@SyberMath]

Thank you! I appreciate that!

[@victorsaldiviacardoch8797]

I need you to make an infinite amount of this videos

[@SyberMath]

I love them, too but I only have a finite number of these! 😁

[@christinayu9598]

Thanks

[@SyberMath]

No problem

[@neuralwarp]

Very nice!

[@SyberMath]

Thanks for the visit

[@carloshuertas4734]

Very good math problem. I got x=1. I thought it was the only solution.

[@DatBoi_TheGudBIAS]

I just looked at the thumbnail and 1 came to my head lol

[@paramjotsingh4682]

Can you bring up a problem on numbers of bases less than 10, in the upcoming videos? It will be great if you do so.

[@SyberMath]

Good idea!

[@paramjotsingh4682]

Thanks. So, please do consider that as well.

[@snejpu2508]

I didn't like the cube root, which demands raising everything to the 6th power. So I made a strange substitution: 2-x=y^3. Now the cube root becomes just y and x-1=1-y^3. So, when I had to deal only with a square root, I squared everything and after some easy manipulations I received 3 solutions for y: 0, 1 and -2. 2-x=y^3, so x=2-y^3. So the solutions for y are 2, 1 and 10. All of them are valid, because they are greater or equal to 1. : )

[@SyberMath]

Thanks for the comment! Would it work if we raised both sides to the 6th power?

[@SyberMath]

Or should we isolate the cube root and then raise both sides to the 6th or 3rd power? 🤔😎😁

[@snejpu2508]

@@SyberMath Probably not. After raising to the 6th power you will still get square roots to 3rd or 5th power, so it wouldn't stop there. You'd have to raise it further... unless there were some really lucky simplifications. But without any substitutions it's probably impossible to solve this.

[@jensonjoseph6296]

@@SyberMath Nope, you would need to isolate then raise both sides to 3rd power, isolate again and square. Directly raising L.H.S. to 6th power using binomial expansion is a nightmare!!!

[@tonyhaddad1394]

Yes radicul are awesome and all math eather !!!!!!!!

[@SyberMath]

Yes, they are! 😍

[@freddyfozzyfilms2688]

x = k^2 + 1

[@christopherrice4360]

nice math problem choice

[@SyberMath]

Glad you think so!

[@myphz1747]

Great explanation! May I ask you which software did you use? Is it available on PC?

[@SyberMath]

Thanks! Notability. It's not available for windows afaik

[@mohammedal-haddad2652]

Awesome

[@SyberMath]

Thank you!

[@marienbad2]

Nice, but what is all this Zee stuff? It's zed! ZED! Also could you do a video on where the idea of using substitutions in equations came from? Who did this first, and what were they trying to solve?

[@SyberMath]

Zee is American, zed is sooo British!!! 😂

[@SyberMath]

This is a good video idea! Thanks!

[@brentwilson6692]

In America, we need to learn our A, Bed, Ceds.

[@SyberMath]

😁

[@umamaheswari7446]

It's mental sum Ans: x = 1

[@SyberMath]

Why?

[@hattorikanzo2793]

When I saw this I guessed som trivial things like 2, 1 and then I thought of a number that would reduce the square root so i guessed 10.

[@SyberMath]

Good guess!

[@Teamstudy4595]

X=1

[@SyberMath]

...and x=2?

[@Teamstudy4595]

yeah Sir that's correct !

[@prakashchander335]

By putting x=1 we find first answer

[@SyberMath]

Nice. Are there other solutions?

[@deepjyoti5610]

Nycc vedio

[@SyberMath]

Thanks!

[@omaral-aghbari681]

My IQ after the video be like:↗️↗️↗️↗️↗️

[@SyberMath]

Great!

[@MrALI4444]

Put u^2=x-1 then 2-x = 1-u^2 and so on ...............

[@SyberMath]

Nice

[@carloshuertas4734]

I understand that the values of x are 1,2, and 10. I got x=1 in my head.

[@SyberMath]

Nice!

[@Arif-ic7tt]

Hello, may I ask where you’re from because your accent sounds like Turkish accent and I am Turkish so I just wanted to know

[@SyberMath]

Merhaba! 😁

[@Arif-ic7tt]

@@SyberMath Merhabalar hocam 😄

[@user-dz6pi4sm8u]

Cool

[@SyberMath]

Thanks!

[@kurtlichtenstein2325]

Good video. Again, I thought I smelled gold, but no.

[@SyberMath]

Sorry! 😁

[@SyberMath]

Gold rush! 😁

[@AshrafAli-qn3gb]

I can only pee

[@aashsyed1277]

What do you mean?

[@SebastienPatriote]

I mean... It's nice to have a technique but x=1 was so obvious here.

[@SyberMath]

Yep!

[@johnnath4137]

These questions generally mean “find ALL the solutions”. Nothing wrong with spotting one or more solutions or even all of them, but having done so you then have to prove rigorously that there aren’t any more.

[@mounirbenjnane6988]

10 is not a solution. But good at all

[@SyberMath]

Thank you! 10 is a solution

[@aashsyed1277]

@@SyberMath yeah!

[@raffaelevalente7811]

I set $2-x=z^3$ that means $x=2-z^3$ and the solution was faster

[@SyberMath]

Cool!

[@otakurocklee]

Very nice!

[@SyberMath]

Thanks!