Solving A Radical Equation Using a Radical Method (maybe)

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[albertmcchan]

Multiply terms by its conjugate, and let y=1/x:
4*(√(1+y)-1) - (1+√(1-y)) = 3y
-√(1-y) = 3*(1+y) - 4*√(1+y) + 2 = 3*(√(1+y) - 2/3)^2 + 2/3
We have LHS ≤ 0, RHS ≥ 2/3, thus y = 1/x is complex.

[Vladimir_Pavlov]

It would be necessary to find the domain of definition of functions in the original equation: IxI≥1.
For x≤-1, the equation that appears at the moment 3: 37 has a left part greater than zero, and a right part less than zero. So there are no solutions in this area. For x≥1, both parts of the equation are positive. But it is easy to see that 4*sqrt(x^2+x) < 5*x+3 and the equation also has no solution in this area.

[kaylashafira]

How did you get x^2-(x^2+x) and x^2-(x^2-x). I am still confused. Thanks

[manojsurya]

The logic that u used at last was great, I didn't see that the quadratic expression is always>0😁

[MathElite]

Oooh this video is gonna be radical in how you solve it

[siddharthabhattacharya]

Can you please take up Greatest integer function floor function ceiling function etc

[sktqmld]

Я не понял с самого начала. Как поменялся знак перед x, во второй дроби. Откуда разность квадратов? И что получилось в числителе первой дроби после привидения к общему знаменателю? Задача решена неправильно!!!

[yoav]

By some manipulations on this equation you get
If x bigger than 1 or smaller than -1 all the expressions are real and 5x+sqrt(x^2-x)+3 is always bigger than 4sqrt(x^2+x).if x between 0 and 1 the left side is complex and the wright is real and if x between 0 and -1 the left side is real and the wright side is complex so there are no solutions at all

[sujanshankarbhowmick]

For the summation where n>1, how many +ve integral solutions for Z are possible?
Dare to do it. It's tough.

[diogenissiganos]

240x^3 + 433x^2 + 270x + 81 = 0

Me after trying to solve this for half an hour only to find out the solution was extraneous:

[yoav]

Intresting .how to show there are ko comples solutions too?

[emanuellandeholm]

I tried the subst t^2 = x^2 + x and u^2 = x^2 - x. x is then (t^2 - u^2)/2. I didn't finish my work (I'm lazy, ok!), but it looked doable.

[jarikosonen]

The last part is most important conclusion..
It looks wolfram alpha couldn't find imaginary solutions either existing. They do not exist?

[carloshuertas]

Another great explanation, SyberMath! It was hard. It ended up with no real solutions.

[epimaths]

Phương trình giải bằng nhân liên hợp. Cảm ơn nhiều.

[hotlatte]

What a plot twist! The last film i have seen like this could be “the shutter island”.

[kuokenwei]

Before you uncovered the solution, I used the same way (rationalize them separately) and found that no roots for the equation. But I just don’t believe it and doubt for my process is correct or not. Think too much again😂.

[rssl]

Btw I understand that both numbers can not be 0 but why can 8x2+15x-9 be -8 times sqrt(x4-x2)
I mean why doesn’t this form have solution ( when it’s non 0 )

[josephsilver]

The equation established at the end of the video shows the quadratic and the last term with the radical. Subtracting the radical term from both sides will make the sign of that term negative meaning the radical term will always be <=0. Since the quadratic expression must be equal to this value it too must be <=0. Since as you pointed, the quadratic value corresponds to a parabola with its minimum point above the x axis - this can never be true - thus there are no real solutions.

[stevenwilson]

This equation has 2 solutions: x = -1, and x = 9/16. You can validate this by graphing (use desmos or your fav graphing app/website)

4/(x + sqrt(x^2 + x) - 1/(x - sqrt(x^2 + x) - 3/x