Can you solve this 𝐏𝐎𝐏𝐔𝐋𝐀𝐑 Equation ? 

Kaise kiya ?

Practice bro 💀

@@Prashant5952_ 🤧

maine bhi

@@khuranapranshu by just hit and trial I also did it, x=8, because 9 is a perfect square

Solution please

√(1+√(1+x))=x^1/3 Squaring on both sides - 1+√(1+x)=x⅔ √1+x=x⅔-1 Squaring on both sides - 1+x=(x⅔)²+1-2x⅔ x^(4/3)-2x⅔-x=0 x⅔(x⅔-2-x^1/3)=0 So x⅔=0 Which meant x=0 Or x⅔-x^1/3-2=0 Let y=x^1/3 So our above equation becomes y²-y-2=0 Solving this we get - y=2 , y=-1 or x^1/3=2, -1 Cubing on both sides, we get- x=8,-1 So total solutions we have- x= 8(rejecting -1 and 8)

thank you​@@MasterPython84

But x can't be -ve

​@@researchsiiopyaa

Sir please demonstrate the solution

Your class i need help

Olympiad isn't class specific! What are you talking about?

@@PriyaSuman-i3c bro i want to start studying 11 class maths or olympiad one like number theory as I want to do jee i am in 10th

@@shubhstudydiaries if you wanna prepare jee and you have no Olympiad background then just do class 11th math first then do some geo,combi and nt (ioqm level)

K=1/4?

No- answer is K=3..

@@Aim92 bro yeh y= (ln(x^ln(x)))^lnx yeh hai na function?

@@Aim92 bro recheck kar lia and k ki value 1/2 aarahi hai Kaha se liya yeh question?

@@jiviteshsahni2085 PW mein AITS hua tha uss mein se liya hai Q...

No dear, When you write √1 it means "1" So, x²=1 does not mean x=√1, it actually means x=+‐ √1 (read, "plus minus root one")

@@TamalRayMahapatra Thanks 🙏

Too easy 😊

This type of Qs is generally done by hit and trial and by this u will get 2 but if you want formal method I would say to plot the exponential and linear functions separately and see how many intersections u r getting no other method Ig

​@@JEE2026IIT r u frm 10th class ??

Use Lambert W function to solve such Equations or use Numerical Methods.

How

Nahi kyoki x tending to very very small value is not given

because root(x^2) = |x|.. not +- x.. ye ek bohot common misconception hai.. hum kabhi kabhi root(9) ko +- 3 likh dete hain.. wo galat hai.. root(9) is 3.

@@subarnodatta ok bro thanks!!!

Thank you for 9.66k subscriber's 😅😅

It's "subscribers" without the apostrophe

Method me hit and trial likhega 😂

Har jagah hit & trial 😅😅

Yea I tried and got sec(thetha)= -3 therefore tan²(thetha) = 8 =x I did it by myself and felt happiness so much

I think it will be complicated bro because RHS will become ^2/3

@@souryajitghosh9341 sir power6 kardo LHS and RHS ki then substitute x=tan²(thetha)

Me kasie de dakta ho bata bro😢

Congratulations bro🎉 phod ke aana

Best of luck bro 🎉 phod ker aana

Bhai root hamesha positive mein khulta hai By definition

STUDY

How? I did everything correctly but sir has -1

No, something is wrong Because at first I also got the same value

In the initial equation, LHS is positive , so even RHS must be positive So, (x)^1/3 >0 , x>0 (even x=0 doesn't satisfy the equation) After squaring on both sides, you get (1+x)^1/2 = (x)^1/3 -1 LHS IS POSITIVE, SO EVEN RHS SHOULD BE POSITIVE, x^1/3>1, so x>1 Therefore, we have to reject 0,1 or you can directly put 0 and 1 in the equation, the equation will not be satisfied