Solving a Radical Equation with Unlike Roots

This video is about solving a complicated radical equation
If you need to post a picture of your solution or idea:
#ChallengingMathProblems #RadicalEquations

EXPLORE:

[agnibeshbasu]

nice! bring more videos like this and olympiad problems.

[maxbow-arrow]

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

[xn____]

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

[nerusfuho]

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

[osmanfb]

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.

[elkincampos]

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

Thus u=0, u=1 or u=3 that is x=u^2+1=1, 2 or 10.

[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.

[ulrichkaiser]

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 }

[misterdubity]

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.

[jeromemalenfant]

???? 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.

[AnthonySpinelli-fevn]

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.

[chhromms.]

easy to understand and great explanation.
perfect

[davidseed]

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

[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.

[victorsaldiviacardoch]

I need you to make an infinite amount of this videos

[quajutsu]

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

[DatBoi_TheGudBIAS]

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

[aliasgharheidaritabar]

Wonderful.you keep me busy with your videos.

[carloshuertas]

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

[marienbad]

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?