A challenging rational equation

This video is about an rational equation with a very interesting solution
If you want to share a picture of your solution/ideas, send me a tweet at @SyberMath and comment here with the link to the tweet.
#ChallengingMathProblems #RationalEquations

EXPLORE:

[SyberMath]

This is a rational equation that is solved by using some algebraic tricks. And with this video we reached 1K subs! Thanks for all the support! 😁

[kaslircribs]

You have a very nice way of solving a rational equation, esp. the restrictions (inequalities) and you have a good sense of humor. Thank you again sir for a great lesson.

[leonhardeuler]

I love this problem. The solution method is simply amazing!

[mangeshbhanage]

If you look at rhs = 7/8. Denominatior is 8 = 2^3
I just assumed that (x+1)^3 =8 which gives us x+1 =2
X =1
If we substitute x=1 the expression becomes

1/1^3 - 1/2^3= 1-1/8 = 8-1/8 = 7/8

[justabunga]

That was a good one. Here is the long way of doing this algebraically, which is finding the LCD, which is 8x^3(x+1)^3. If we multiply the equation by that LCD, we get 8(x+1)^3-8x^3=7x^3(x+1)^3. For simplification, we get Setting the equation to 0 gives us Using rational roots theorem, we know that x=1 is a solution. Using the factored form, we get Therefore, x=1 and x=-2. The other factored form has no real solutions.

[beautifulmindinpuzzles]

Alternate Solution:
Combining the LHS we soon reach to:
[3(x^2 + x)+1]/(x^2+x)^3 = 7/8
Let x^2 + x = y, we have the cubic equation 7y^3 - 24y -8 =0 whose real solution is y = 2. Then x = 1 and x = -2

[khundeejai]

It obvious that 1-1/8=7/8 but it can be written -1/8+1=7/8 so at least 1 and -2 are solutions.

[lsys]

I enjoyed this video very much. I thought about it a lot and solved it over and over to fully understand it. I hope to have learned these useful algebraic tricks.

[tomasstride]

The thing is solving problems is often determined by what direction is chosen at the start. I looked at this problem without starting the video and thought I don't want to end up with some sort of sixth order polynomial to solve. So I thought I would try to symmetrize the problem by putting x=y-1/2. It sort of worked . You do end up with a cubic (in y**2-1/4) which is probably not as nice as the one you ended up with but did give the right result. My approach was in the end not as good as yours but it's hard to see this at the start.

[rafael]

I am watching all your videos. Congratulations for your work. The selection of the problems and the resolutions are very "OK" (sorry for mi English, I am Spanish)

[mxsjncv]

obviously knowing how to solve it like you do, is great. I tried this way: drawing y=1/x^3 is easy. drawing the same function that was "mooved" one unit to the left is easy. Now we see that, for example for x>0 the vertical distance between these two functions is decreasing, as we go to the right. So we try x=1, x=2 .... Same idea in the left side. In this case it "works".

[navyabhatia]

I checked it by taking the value of x=1 and it works so I divided the whole equation as x-1 will be the factor and then for the equation after division we get it works for x=-2 so x+2 as a factor and then for other equation after division no roots can be possible so in this way I had done this.By the way love your videos sir💖 keep growing and uploading 👍.Love from India 🇮🇳.

[prabhudasmandal]

Your selection of problems and their solutions are mind blowing, no doubt. This equn is easily solved and can be kept in mind by taking x+1=t and x=t_1 resulting in 3t^2_3t+1)/(t^2_t^3), again taking (t^2_t)=m it is finally (3m+1)/m^3=7/8 which yields the result.

[kinshuksinghania]

It is 1 - 1/8 if you look closely!!. Hence x=1 is a solution. Other solutions can determined after factoring out (x-1)

[wisdomokoro]

I expanded and let the denominator be (x^2+x)^3 and simplifying I got a nice cubic in the variable x^2+x and surely, one of which by trial and error gave me 2 easily (a standard method by the way of solving cubics🤫🤫. And the other 2 were complex roots🤔🤔

[robyzr]

hello dear teacher. Maybe I could wrong after many years ..lol but at about minute 7.00 when u find the polinomial u^3 + 6u^2 - 7 = 0, by Ruffini mehods I find (u - 1) ( u^2 + 7u + 7 ) and the final solutions are: u = 1 ; u = -1 ; u = - 6 . Last solution give me anyway complex solution

[thornadotrigger]

you are super intelligent at math, .i love your videos

[dneary]

I rearranged the original equation by noticing that 1/x^3>1/(x+1)^3 for x>0 and 7/8=1-1/8 is also a difference of cubes.
Rearranging: 1/x^3 - 1 = 1/(x+1)^3 - 1/2^3 - using the difference of cubes formula, we see that the LHS is (1-x)/x (1+1/x + 1/x^2) and the RHS is (1-x)/2(x+1)(1/4 + 1/2(x+1) + 1/(x+1)^2) = so either 1-x = 0 or 1/x(1+1/x+1/x^2) = 1/2(x+1)(1/4 + 1/2(x+1) + 1/(x+1)^2 but we can compare the terms pairwise to prove that 1/x(1+1/x+1/x^2) > 1/2(x+1)(1/4 + 1/2(x+1) + 1/(x+1)^2 for all x>0 (and we know that x>0 because 1/x^3-1/(x+1)^3 = 7/8 > 0 ).

[someshkaashyap]

Riddle me this,

Simplify lhs and you have (x)*(x+1) as denominator

The denominator is 8

So x(x+1)=2
X=1...

What's so difficult?

[epimaths]

Bài toán phương trình được giải bằng phương pháp sử dụng hai ẩn phụ. Cảm ơn.