erros x = 1 is not a root and 2nd loop solved is wrongly done. . . To master your problem-solving skills up to JEE Advanced join our course jeesimplified.com/set-of-60 . Send us the hardest question solved by you forms.gle/HZxgUAKdWV1Pywgb8
Pehle hi bata du fekne me muze koi interest nahi hai ... jee main 2024 30 jan shift 1 99.91pr hai mera ......Bolte hue bura to lag rha hai bavjood ki me khood bohot achha feel karunga agar koi meri baat samjega ... i know it is difficult to provide such amazing content free on yt but in previous time bhai aapne vo chiz provide karai thi .. aap apne set of 60 ke behtarine sawal late the yaha pe .. magar ab nahi late .. jo bhi reason ho magar aap wo content nahi to us level ka content plz provide karate rahie ....ha ese method kaam ke hai magar pura paper method wale ese qn ka to sirf calculation me use karenge na ki qn ki thought process me jo aap purane qn me build karvate the 😢😊😊
I solved it in like 3 mins with a completely different approach: sqrt (5-x)=5-x^2=y (let) Then, the two equations can become: y=sqrt (5-x) or x+y^2=5 And, y= 5-x^2 or x^2+y=5 Hence, equating the two equations, x^2+y= x+y^2 x^2-y^2 - (x-y)=0 (X+y)(x-y) - (x-y)=0 (X-y)(x+y-1)=0 This gives either x=y or y=1-x (i) x=y in first eqn: x^2+x=0-> solve to get two solutions (ii) y=1-x-> x^2+(1-x)=5 or x^2-x-4=0-> solve to get two more solutions.
I was able to solve the Q with same approach because of the hint given in the thumbnail 😌 And to eliminate extra values we can put them in original equation and discard the values at which we are getting negativite value inside the sq. root .
Me jnv etawah se hu or jnv ke entrance exam me esy questions √5-√5-√5-........... infinity ho to usko √4a+1 +-1 /2 se krte to x=√5-√5-√5....... ♾️ X ki value √21-1 / 2 lete or yh iska answer ho jata oor apka answer bhi yhi a rha hai or mera bhi .....🥰 Thank you bhaiya nya tarika btane ke liye pr agr mere method me glti ho to pls correct kr dijiye ga 😊
Bhaiya aapke thumbnails se Aisa kyo lagta hai ki aap iitians ko over glorify kar rahe ra ho.. isn't it vague as an educator on moral grounds? I hope you understand, it's complicated. I am just concerned about those insecure students who have an exposure to glorification of IITians or similar things like things happening in Kota would be an example of this toxicity. 😢 I think we all must focus on excelling not being a gimmick with a prior "tag".
Another way i solved it in: Let y=√(5-x). Let us find its inverse function. We see that the inverse of √(5-x) is 5-x^2 !!! So the solutions to √(5-x)=5-x^2 will be on the line y=x (since inverses are reflections about y=x, their intersections will lie on y=x) hence we can equate 5-x^2=x giving us x^2+x-5=0 giving us x= (-1±√(21))/2
Bro there are four solutions in my opinion (because I haven't seen the video) x = (-1±√(21))/2 and (1±√(17))/2. I did it taking 5 to be variable and x to be constant then I solved it.
10:40 the value can be substitued in the original eqn which was 5-x= (5-x^2)^2 , we would get the value of x that are the soln , and also if we sketch graphs of root(5-x) and 5-x^2 , we will get only two intersection thus only two ans.....
@@shivankshrivastav damn its my dream to work at amd, although I am still confused between cse and hardware engg, tbh it will be decided by what rank I get
I solved it as an intersection of 2 parabolas. y² = 5-x and (only top part) and x²=5-y. Now if we subtract the two we will get y²-x²=y-x so either y+x-1 = 0 or y-x=0. Now we can put these cases and solve for 4 roots then check for which values y is +ve (as we are only considering the upper segment of y²=5-x)
it is also worth noting that the parabolas are inverse of each other and then by graph we can easily notice that y+x-1=0 or x=y and then we can proceed with your solution... it felt amazing coming up with this solution
After I got x²+x-5 as one factor, in the quartic equation i just divided it by the factor and got both quadratics and solved them😅 , couldn't think of your method before the video unfortunately, however we got 2 extra roots by squaring on both sides but the square root doesn't give both ± so -√5≤ x ≤√5 , I removed extraneous solutions by putting it in the orginal equation 👍
This question can become more easy tell me my method is right or wrong according to the method √ 5 - x is equal to 5 minus x square root 5 is equal to 5 minus x square + x here 5 minus x square + X is and quadratic equation find the value of x using the quadratic formula we got the answer that x is equal to minus 1 + root 21 upon 2 which is the answer
Bolte hue bura to lag rha hai bavjood ki me khood bohot achha feel karunga agar koi meri baat samjega ... i know it is difficult to provide such amazing content free on yt but in previous time bhai aapne vo chiz provide karai thi .. aap apne set of 60 ke behtarine sawal late the yaha pe .. magar ab nahi late .. jo bhi reason ho magar aap wo content nahi to us level ka content plz provide karate rahie ....ha ese method kaam ke hai magar pura paper method wale ese qn ka to sirf calculation me use karenge na ki qn ki thought process me jo aap purane qn me build karvate the 😢😊😊
Solve for x over the real numbers: sqrt(-x + 5) - (-x^2 + 5) = 0 sqrt(-x + 5) - (-x^2 + 5) = -5 + x^2 + sqrt(-x + 5): -5 + x^2 + sqrt(-x + 5) = 0 Subtract x^2 - 5 from both sides: sqrt(-x + 5) = -x^2 + 5 Raise both sides to the power of two: -x + 5 = (-x^2 + 5)^2 Expand out terms of the right hand side: -x + 5 = x^4 - 10 x^2 + 25 Subtract x^4 - 10 x^2 + 25 from both sides: -x^4 + 10 x^2 - x - 20 = 0 The left hand side factors into a product with three terms: -(x^2 - x - 4) (x^2 + x - 5) = 0 Multiply both sides by -1: (x^2 - x - 4) (x^2 + x - 5) = 0 Split into two equations: x^2 - x - 4 = 0 or x^2 + x - 5 = 0 Add 4 to both sides: x^2 - x = 4 or x^2 + x - 5 = 0 Add 1/4 to both sides: x^2 - x + 1/4 = 17/4 or x^2 + x - 5 = 0 Write the left hand side as a square: (x - 1/2)^2 = 17/4 or x^2 + x - 5 = 0 Take the square root of both sides: x - 1/2 = sqrt(17)/2 or x - 1/2 = -sqrt(17)/2 or x^2 + x - 5 = 0 Add 1/2 to both sides: x = 1/2 + sqrt(17)/2 or x - 1/2 = -sqrt(17)/2 or x^2 + x - 5 = 0 Add 1/2 to both sides: x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x^2 + x - 5 = 0 Add 5 to both sides: x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x^2 + x = 5 Add 1/4 to both sides: x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x^2 + x + 1/4 = 21/4 Write the left hand side as a square: x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or (x + 1/2)^2 = 21/4 Take the square root of both sides: x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x + 1/2 = sqrt(21)/2 or x + 1/2 = -sqrt(21)/2 Subtract 1/2 from both sides: x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x = sqrt(21)/2 - 1/2 or x + 1/2 = -sqrt(21)/2 Subtract 1/2 from both sides: x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x = sqrt(21)/2 - 1/2 or x = -1/2 - sqrt(21)/2 sqrt(-x + 5) - (-x^2 + 5) ⇒ sqrt(5 - (1/2 - sqrt(17)/2)) - (5 - (1/2 - sqrt(17)/2)^2) = 1/2 (-1 - sqrt(17) + sqrt(2 (sqrt(17) + 9))) ≈ 0: So this solution is correct sqrt(-x + 5) - (-x^2 + 5) ⇒ sqrt(5 - (sqrt(17)/2 + 1/2)) - (5 - (sqrt(17)/2 + 1/2)^2) = -5 + sqrt(9/2 - (sqrt(17))/2) + (sqrt(17)/2 + 1/2)^2 ≈ 3.12311: So this solution is incorrect sqrt(-x + 5) - (-x^2 + 5) ⇒ sqrt(5 - (-sqrt(21)/2 - 1/2)) - (5 - (-sqrt(21)/2 - 1/2)^2) = 1/2 (1 + sqrt(21) + sqrt(2 (sqrt(21) + 11))) ≈ 5.58258: So this solution is incorrect sqrt(-x + 5) - (-x^2 + 5) ⇒ sqrt(5 - (sqrt(21)/2 - 1/2)) - (5 - (sqrt(21)/2 - 1/2)^2) = -5 + sqrt(11/2 - (sqrt(21))/2) + (sqrt(21)/2 - 1/2)^2 ≈ 0: So this solution is correct The solutions are: Answer: | | x = 1/2 - sqrt(17)/2 or x = sqrt(21)/2 - 1/2
This question was actually designed/made to solve after realizing that the given is one and only f(x) = f¯¹(x) and we have to solve f(x)=x to get to our solution... But still there's a catch here, after solving via this method we'll end up in only one root, so for the other root we'll have to observe some critical points from graph of both fns and then we will get to know that the other root (-ve one) will actually lie on a line with slope - 1 and it's y intercept will come out to be the sum of x and y coordinates... After this we can solve directly by equating both the eqns and we will easily get the sum of coordinates of x, y to be 1 hence we will get our line and after that is our standard approach.... Also for rejecting 2 solns in your soln bhaiya we will use graphical approach and there we will be having 2 roots one negative one positive Edit : I solved in about 5mins Also I knew your approach bhaiya, but didn't went through it because it was long(for me atleast)
@@Mathlover_1729 I also thought the same before my method but I then I realized that no we cannot solve it by assigning x to trignomentric ratio... Since sin cos cosec and sec ratios are either bounded or does not have real range, because we don't know whether x will be having solutions less than 1 or greater than 1 same goes for - 1.... We can surely put x to some tan@ or cot@ and solve since then it will solve for all real x... I hope I was able to clarify it
My method, (5-x)^1/2 = 5 - x + x -x^2 Now taking 5 - x on left side and taking( 5-x)^1/2 as common (5-x)^1/2 [ 1 - (5-x)^1/2] = x [ 1 - x] Now if we compare both sides we can equate (5-x)^1/2 with x And then after squaring both sides we get the required quadratic which is x^2 + x - 5 = 0.......and applying quadratic formula to get roots
i was able to think by substituting y assuming y = sq.rt of 5-x. Thinking to relate it as parabola and Quadratic graph the only essence that the graph rotated. So able to guess x=y.
Thanks bhaiya for taking my suggested question. Ye question Maine ek bade achhe channel se liya tha, jiska naam hai "BlackPenRedPen". Us channel pe high school maths se high level maths ke achhe question mil jaate hain. Ye mujhe vahine se mila tha. I'll recommend everyone to checkout that channel also👍
i didn't take this as a quadratic of 5, ill explain it in brief here, i made the quadraric in (1/2 + x ^2) added and subtractited x^2 and 1/4 , other quadratic was made as a result of it, i did a^2-b^2 = a-b)(a+b) and then boom, same two last quadratics as you, this took 3-4 mins of thinking
one of the easiest approcach and in very less steps is: root(5-x) = 5- x^2 rearranging x^2 = 5 - root(5-x) taking root both sides x = root( 5 - root(5-x)) and we can replace x in RHS by the value of x from the LHS, so then x = root(5-root(5-root(5-x))) and repeat so on, and if we look reverse way then, x= root(5-x), then squaring both sides then we get x^2 = 5-x, and hence we have solved this SO CALLED AMAZING EQUATION by PATTERN
But this step is a bit risky because you are not sure if the D (b²-4ac) will be a perfect square and if it wouldn't have been a perfect square it would have been more difficult that way.
Ye achhi community h doston...time agr h 2 saal acche se to follow krna..agr last 3 4 months me ho to follow pyqs only and modules of your coaching.. AIR 1789 (JEE 2022) here
Yahaan (x) ko subject rakh hi nahi rahe hain (hence the title 'value of 5')... Ye demonstration Mr. Aman Malik ne bhi kuch kuch videos mein diya hua hai. Of course, no practice means I'd forget eventually. Lekin memory refresh ho gai 7:20 par! PS: I didn't take the JEE, Mr.Simplified... I just like math and your channel is really interesting.
Change your channel name to something related to maths.. that way channel will go in a different bigger numbers... Take maths conspiracies and all... Thank me later!
Apne under root me (2x²-1)²-4(x⁴+x) liya tha...par aap uss (2x²-1)² ki jaga (2x²+1)² Lena tha ....tab solution pakka hai🔥🔥🔥...well done and good job sir and Aditya too🎉❤❤
5-x negative ni hona chaiye To, x is less than or equal to 5 Ye ek condition aagyi Dusri condition h ki 5-x^2 negative ni aana chaiye Mtlb x^2 is less than or equal to 5 Put krenge to do values reject ho jaengi
graphical method has always been best. plot both the graphs and since they are inverse of each other, one or more root lies on y=x 5-x^2=y=x (first quadratic)or √(5-x)=y=x
Another method i tried: subtracting x from both sides √(5-x)-x=5-x^2-x rationalizing both sides [√(5-x)-x][√(5-x)+x]/[√(5-x)+x]=5-x^2-x simplifying the numerator [5-x^2-x]/[√(5-x)+x]=5-x^2-x Cancelling out Numerator of LHS and RHS ( noting that we have a solution where 5-x^2-x =0)[quadratic no.1] [√(5-x)+x]=1 bring x on rhs and now you can solve the quadratic no.2
WE can eliminate the values based on the domain of x we get . 5-X is under square root thus x must be less than 0 and 5-x^2 must be more than or equal to 0. So we get x must lie between ..... negative root 5 to root 5 . So we are done.
so first i graphed both the functions to check the number of real solutions, they intersected at 2 point (1 +ve & 1 -ve). Now i started solving the sum, i took the x^2 term to the other side (x^2 = 5 - (5-x)^1/2) now sbs we get x = +- ( 5 - (5-x)^1/2)^1/2 now we plug in the value of x in this equation and get a infinite loop of -ve square root of 5 now assume that to be x sbs get a quadratic and we got the positive root now we can use -part of x [-(5-(5-x)^1/2] to get our other solution in a similar way but this time we will get alternate +ve and -ve square root of five in our loop.
erros
x = 1 is not a root and 2nd loop solved is wrongly done.
.
.
To master your problem-solving skills up to JEE Advanced join our course
jeesimplified.com/set-of-60
.
Send us the hardest question solved by you
forms.gle/HZxgUAKdWV1Pywgb8
Error*
@@puchokaun error likhne me bhi error ho gyi bhai se 😂
😅 yeah I have seen this
@@yuraje4k348 toh kya bhai? hota hai insaan hai wo bhi
Pehle hi bata du fekne me muze koi interest nahi hai ... jee main 2024 30 jan shift 1 99.91pr hai mera ......Bolte hue bura to lag rha hai bavjood ki me khood bohot achha feel karunga agar koi meri baat samjega ... i know it is difficult to provide such amazing content free on yt but in previous time bhai aapne vo chiz provide karai thi .. aap apne set of 60 ke behtarine sawal late the yaha pe .. magar ab nahi late .. jo bhi reason ho magar aap wo content nahi to us level ka content plz provide karate rahie ....ha ese method kaam ke hai magar pura paper method wale ese qn ka to sirf calculation me use karenge na ki qn ki thought process me jo aap purane qn me build karvate the 😢😊😊
I solved it in like 3 mins with a completely different approach:
sqrt (5-x)=5-x^2=y (let)
Then, the two equations can become:
y=sqrt (5-x) or x+y^2=5
And, y= 5-x^2 or x^2+y=5
Hence, equating the two equations,
x^2+y= x+y^2
x^2-y^2 - (x-y)=0
(X+y)(x-y) - (x-y)=0
(X-y)(x+y-1)=0
This gives either x=y or y=1-x
(i) x=y in first eqn: x^2+x=0-> solve to get two solutions
(ii) y=1-x-> x^2+(1-x)=5 or x^2-x-4=0-> solve to get two more solutions.
👍👍👍👍👍
Nice😊
i) =5 ✓ not =0
Copied straight from blackpen red pen 😮
I see a man of culture here
Can you give me the link of video
Haan Bhai, vahine se Diya tha maine😊, khud to itne badiya sawal nahi bana sakta na😅
@@adityaagarwal636hello aap ne hi ye sawal bhaiya ko Diya tha
@@UmG_Editzhaan bhai
x = ✓(5-x)
divide both sides by, ✓(5-x)
x/(✓(5-x)) = 1
rationalize to get,
x(✓(5-x)) /(5 - x) = 1
square both sides to get,
(x^2)(5-x)/(5-x)^2 = 1
(x^2)/(5-x) = 1
x^2 = 5-x (by multiplying 5-x to both sides)
now, get the quadratic equation,
x^2+x-5 = 0
finally solve the quadratic
I was able to solve the Q with same approach because of the hint given in the thumbnail 😌
And to eliminate extra values we can put them in original equation and discard the values at which we are getting negativite value inside the sq. root .
X = 1 kaam kaise kr rha hai ? Check kro equation me x = 1 daalke
Bhai 2 min tak pagal ho raha tha ki x=1 kaam kaise kar raha hai
Mujhe laga mai maths bhul gaya
@@prince-hb8qk same 💀
Exactly
Bhai nasha karna band krde
Same
7:20 quadratic in 5
haaa bhaiya main hi hi 1/ plancks constant sec me banane wala
its plancks constant
Woh kiya hota hai?
😂😂😂🤣@@noexistence88
Plancks constant h ~ 10^-34
1/h~ 10^34
10^34 seconds means 10^26 years !!
Think once....
bhaisab, tum J^-1s^-1 mei banate ho question? Kya logic hai.
7:27 solve by dharacharya and 5 as a root
Me jnv etawah se hu or jnv ke entrance exam me esy questions √5-√5-√5-........... infinity ho to usko √4a+1 +-1 /2 se krte to x=√5-√5-√5....... ♾️ X ki value √21-1 / 2 lete or yh iska answer ho jata oor apka answer bhi yhi a rha hai or mera bhi .....🥰 Thank you bhaiya nya tarika btane ke liye pr agr mere method me glti ho to pls correct kr dijiye ga 😊
anyone who has watched the video how real men solve equations knew this approach
and it was discussed by blackpenredpen as well
8:50 me (2x²-1)² kyu hai (2x²+1)² hoga na due to b ka value
Yeh you're right
Are Ho jati h galti....
bhai new vids kab aaegi
It's a inversely fn concept 5-x^2=x and solvr
I just let x=5-y and put it in original equation then we get a quadratic in y and after solving we get x
We can also convert this to infinite nested root and we can solve that easily
thats what he did
Yeh itne raat ko kyu upload krte ho bhai isse log jldi notification pr click nhi kr paenge yt boost nhi dega vdo ko 6-7 bje shaam ke krte jao jb sb log yt dekhte hai
Bhai raat ko hi sahi hai sab khatam hone ke baad milta hai ....6 7 baje busy rhenge target audience aur fir bhul jaenge baki notifications ke andar
AS bprp said, believe in the power of geometry, considering making a triangle with said thingy
If 5 only root hai to b²-4ac must be 0
Lekin use correct x nhi aayega
7:23 '5' me quadratic form ho gyi, toh quadratic formula lgayenge 5 ki value nikalne ke liye
It is in Allen module function 12 question about no. Of sol
10:20 Surtz negative ke barabar nahi ho sakta
He he 5 me quad bana di caught it at 7:29 maths😮😮
05:05 par plus 20 hoga
7:30 I think it's a quadratic equation in 5 let's see
9:02 mei perfect square kaise bana?
Bhaiya consistentcy tut gyi aapki 69th(🙂) video par
Bas 6 aur bacchi thi
App used?
Bring x square to lhs
badiya question
Vai conventional methods me tu galat kardiya yar
quadratic in x banana hai bprp ka vdo same tha
How the hell,1 is not factor of it
6:50 i got it
Copied from Blackpen and red pen which was uploaded 5 years ago pn his Chanel
Bhaiya aapke thumbnails se Aisa kyo lagta hai ki aap iitians ko over glorify kar rahe ra ho.. isn't it vague as an educator on moral grounds? I hope you understand, it's complicated. I am just concerned about those insecure students who have an exposure to glorification of IITians or similar things like things happening in Kota would be an example of this toxicity. 😢
I think we all must focus on excelling not being a gimmick with a prior "tag".
🤔 I thought it adds on humour to but damn, you are right bro
Will strongly consider this opinion
Acha community h bhoi log....
JEE Adv 2016 rank 5574 here❤
Hello bhaiya can u please guide me I just moved to 12
Aise comment ke nhi hoga ache se guide..bhaiya ka paid mentorship le le @AbhishekRaj
@@iitiandev121 bhaiya wo jyada ho jayega abhi utna afford nhi kr skta
Konse iit m ho bhaiya
@@reviewer3562 Tanishq rajak?
Dhakkan ho kya? Shuru ke 2 mins main hi itni badi galti.... x = 1 kaise solution hai iska? Matlab kuch bhu......
I mentioned the error bro, at times while making the video mind zone out hoojata hai.
Reminds me of that problem in "pair of straight lines", applying the concept of homogenization of a curve, taking 1 as a variable.
Yeah man
Another way i solved it in:
Let y=√(5-x). Let us find its inverse function. We see that the inverse of √(5-x) is 5-x^2 !!!
So the solutions to √(5-x)=5-x^2 will be on the line y=x (since inverses are reflections about y=x, their intersections will lie on y=x) hence we can equate 5-x^2=x giving us x^2+x-5=0 giving us x= (-1±√(21))/2
I did it by this method, but these functions are inverses only for x>0. You'd have to find the other solution by some different method.
I think im only to do like this 😂
Bro there are four solutions in my opinion (because I haven't seen the video) x = (-1±√(21))/2 and (1±√(17))/2.
I did it taking 5 to be variable and x to be constant then I solved it.
This doesn’t always work.
@@ryanrahuelvalentine2879 oh really
X=1 toh root hai hee nahi?Factor kese kar diya uska
10:40 the value can be substitued in the original eqn which was 5-x= (5-x^2)^2 , we would get the value of x that are the soln , and also if we sketch graphs of root(5-x) and 5-x^2 , we will get only two intersection thus only two ans.....
Nostalgia of jee days
jee adv 2018 1331 here
what are u doing now dude
@@lifeisamarathon2098 I'm currently working in AMD as an lithography engineer at TSMC, Taiwan.
@@shivankshrivastav damn its my dream to work at amd, although I am still confused between cse and hardware engg, tbh it will be decided by what rank I get
@@shivankshrivastav 🙌🙌hat's off, wishing for your best.
@@shivankshrivastav abhi allen m marks nhi aa rhe but theory se coaching k sath hu kya me abhi bhi iit bombay crack kr skta hu
I solved it as an intersection of 2 parabolas. y² = 5-x and (only top part) and x²=5-y. Now if we subtract the two we will get y²-x²=y-x so either y+x-1 = 0 or y-x=0. Now we can put these cases and solve for 4 roots then check for which values y is +ve (as we are only considering the upper segment of y²=5-x)
We have to consider this( -✓5,✓5)domain to
Thats correct but must be time consuming
it is also worth noting that the parabolas are inverse of each other and then by graph we can easily notice that y+x-1=0 or x=y and then we can proceed with your solution... it felt amazing coming up with this solution
After I got x²+x-5 as one factor, in the quartic equation i just divided it by the factor and got both quadratics and solved them😅 , couldn't think of your method before the video unfortunately, however we got 2 extra roots by squaring on both sides but the square root doesn't give both ± so -√5≤ x ≤√5 , I removed extraneous solutions by putting it in the orginal equation 👍
7:04 Understood, we have make a quadratic taking 5 as a variable and finding it's value in terms of x.....❤
Are bhai ye Blackpenredpen ne explain kiya h....maine ise 9th me hi solve kiya tha
People laugh at me whenever I have tried some similar approaches like this😢😢
They think I'm just crazy.
Are gajab
Eise sochna bhi ho skaata
Q chota sa hn par Eise karke kai equation mein apply kar sakenge
Thanks Bhaiya ji
One doubt. Agar quadratic 5 me banali humne then sum of roots and product of roots kiske equal hoga?🤔🤨
Equal roots ka case hai dono root 5 hi hai
Easy question, done in 1/1+2+3+...+nth second and he copied this question from blackpanrendpen.
Jo keh rhe h unse 1 baar me solve ho gya khud se
Meanwhile their jee main %ile
This question can become more easy tell me my method is right or wrong according to the method √ 5 - x is equal to 5 minus x square root 5 is equal to 5 minus x square + x here 5 minus x square + X is and quadratic equation find the value of x using the quadratic formula we got the answer that x is equal to minus 1 + root 21 upon 2 which is the answer
I have solved such question on bhannat maths channel in which he habe given cubic so it was eazy for me
2:00 cardan method
Ferrari method
Kya kre fir😂😂😂
Bolte hue bura to lag rha hai bavjood ki me khood bohot achha feel karunga agar koi meri baat samjega ... i know it is difficult to provide such amazing content free on yt but in previous time bhai aapne vo chiz provide karai thi .. aap apne set of 60 ke behtarine sawal late the yaha pe .. magar ab nahi late .. jo bhi reason ho magar aap wo content nahi to us level ka content plz provide karate rahie ....ha ese method kaam ke hai magar pura paper method wale ese qn ka to sirf calculation me use karenge na ki qn ki thought process me jo aap purane qn me build karvate the 😢😊😊
10;19 root ke andar sq. Hai toh mod se khulega aur do case banege ....jinhe solve karte hi ek ek eliminate ho jaenge because of mod ki condition
Solve for x over the real numbers:
sqrt(-x + 5) - (-x^2 + 5) = 0
sqrt(-x + 5) - (-x^2 + 5) = -5 + x^2 + sqrt(-x + 5):
-5 + x^2 + sqrt(-x + 5) = 0
Subtract x^2 - 5 from both sides:
sqrt(-x + 5) = -x^2 + 5
Raise both sides to the power of two:
-x + 5 = (-x^2 + 5)^2
Expand out terms of the right hand side:
-x + 5 = x^4 - 10 x^2 + 25
Subtract x^4 - 10 x^2 + 25 from both sides:
-x^4 + 10 x^2 - x - 20 = 0
The left hand side factors into a product with three terms:
-(x^2 - x - 4) (x^2 + x - 5) = 0
Multiply both sides by -1:
(x^2 - x - 4) (x^2 + x - 5) = 0
Split into two equations:
x^2 - x - 4 = 0 or x^2 + x - 5 = 0
Add 4 to both sides:
x^2 - x = 4 or x^2 + x - 5 = 0
Add 1/4 to both sides:
x^2 - x + 1/4 = 17/4 or x^2 + x - 5 = 0
Write the left hand side as a square:
(x - 1/2)^2 = 17/4 or x^2 + x - 5 = 0
Take the square root of both sides:
x - 1/2 = sqrt(17)/2 or x - 1/2 = -sqrt(17)/2 or x^2 + x - 5 = 0
Add 1/2 to both sides:
x = 1/2 + sqrt(17)/2 or x - 1/2 = -sqrt(17)/2 or x^2 + x - 5 = 0
Add 1/2 to both sides:
x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x^2 + x - 5 = 0
Add 5 to both sides:
x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x^2 + x = 5
Add 1/4 to both sides:
x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x^2 + x + 1/4 = 21/4
Write the left hand side as a square:
x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or (x + 1/2)^2 = 21/4
Take the square root of both sides:
x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x + 1/2 = sqrt(21)/2 or x + 1/2 = -sqrt(21)/2
Subtract 1/2 from both sides:
x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x = sqrt(21)/2 - 1/2 or x + 1/2 = -sqrt(21)/2
Subtract 1/2 from both sides:
x = 1/2 + sqrt(17)/2 or x = 1/2 - sqrt(17)/2 or x = sqrt(21)/2 - 1/2 or x = -1/2 - sqrt(21)/2
sqrt(-x + 5) - (-x^2 + 5) ⇒ sqrt(5 - (1/2 - sqrt(17)/2)) - (5 - (1/2 - sqrt(17)/2)^2) = 1/2 (-1 - sqrt(17) + sqrt(2 (sqrt(17) + 9))) ≈ 0:
So this solution is correct
sqrt(-x + 5) - (-x^2 + 5) ⇒ sqrt(5 - (sqrt(17)/2 + 1/2)) - (5 - (sqrt(17)/2 + 1/2)^2) = -5 + sqrt(9/2 - (sqrt(17))/2) + (sqrt(17)/2 + 1/2)^2 ≈ 3.12311:
So this solution is incorrect
sqrt(-x + 5) - (-x^2 + 5) ⇒ sqrt(5 - (-sqrt(21)/2 - 1/2)) - (5 - (-sqrt(21)/2 - 1/2)^2) = 1/2 (1 + sqrt(21) + sqrt(2 (sqrt(21) + 11))) ≈ 5.58258:
So this solution is incorrect
sqrt(-x + 5) - (-x^2 + 5) ⇒ sqrt(5 - (sqrt(21)/2 - 1/2)) - (5 - (sqrt(21)/2 - 1/2)^2) = -5 + sqrt(11/2 - (sqrt(21))/2) + (sqrt(21)/2 - 1/2)^2 ≈ 0:
So this solution is correct
The solutions are:
Answer: |
| x = 1/2 - sqrt(17)/2 or x = sqrt(21)/2 - 1/2
😮😮😮😮😮😮😮😮😮🎉🎉🎉🎉🎉❤❤❤❤❤❤❤❤❤😮😮😮😮😮😮❤❤❤
This question was actually designed/made to solve after realizing that the given is one and only f(x) = f¯¹(x) and we have to solve f(x)=x to get to our solution... But still there's a catch here, after solving via this method we'll end up in only one root, so for the other root we'll have to observe some critical points from graph of both fns and then we will get to know that the other root (-ve one) will actually lie on a line with slope - 1 and it's y intercept will come out to be the sum of x and y coordinates... After this we can solve directly by equating both the eqns and we will easily get the sum of coordinates of x, y to be 1 hence we will get our line and after that is our standard approach.... Also for rejecting 2 solns in your soln bhaiya we will use graphical approach and there we will be having 2 roots one negative one positive
Edit : I solved in about 5mins
Also I knew your approach bhaiya, but didn't went through it because it was long(for me atleast)
Nice approach bro... But can we not solve by putting x=5cosΩ Ω€[-π/2,π/2] ... My process is short
@@Mathlover_1729 I also thought the same before my method but I then I realized that no we cannot solve it by assigning x to trignomentric ratio... Since sin cos cosec and sec ratios are either bounded or does not have real range, because we don't know whether x will be having solutions less than 1 or greater than 1 same goes for - 1.... We can surely put x to some tan@ or cot@ and solve since then it will solve for all real x... I hope I was able to clarify it
After finding 1st two roots by putting y=x we can square original equation and divide it by eqn having two roots we got earlier
@@ParthBnsl-iitis yes bro,this is the problem
Ferrari method se hogaya use mostly 4 degree ho jati hai solve
Bhaiya, y=√5-x and y=5-x² inverse functions hai.Then y=x pe unka root lie karega.
only for x>0. Ek solution aajayega, doosra nahi.
eliminate kiye
using 1st equation
5-x^2>=0
Actually, root(17) +1/2 won’t be the solutions as they don’t satisfy the original equation. They are extra solutions obtained from squaring.
My method,
(5-x)^1/2 = 5 - x + x -x^2
Now taking 5 - x on left side and taking( 5-x)^1/2 as common
(5-x)^1/2 [ 1 - (5-x)^1/2] = x [ 1 - x]
Now if we compare both sides we can equate (5-x)^1/2 with x
And then after squaring both sides we get the required quadratic which is
x^2 + x - 5 = 0.......and applying quadratic formula to get roots
I did with same method
10:41 cases orignal eq se eliminate ho jayenge since x² is less than 5 therefore x would be approx less than 2.23🙂
Dimaag phat gya ye dekh ke
Bhai , baaki videos nhi kroge upload?
mai jo blackpenredpen dekhta ru roj ☠☠
i was able to think by substituting y assuming y = sq.rt of 5-x.
Thinking to relate it as parabola and Quadratic graph the only essence that the graph rotated.
So able to guess x=y.
7:10 pe pata hai 5=shree dharacharya expression in x likhoge as given in title of video , i used to do shxt like this alot good its gettin traction
1:28 apply rational root theorem agar sare zeros irrational and complex nhi h to work krega
Thanks bhaiya for taking my suggested question. Ye question Maine ek bade achhe channel se liya tha, jiska naam hai "BlackPenRedPen". Us channel pe high school maths se high level maths ke achhe question mil jaate hain. Ye mujhe vahine se mila tha. I'll recommend everyone to checkout that channel also👍
Lovely sol. bro
Congrats on completing the challenge.💯
Bhai mujhe maths me 98 percentile aai thi
To mai naachu kya😂
Ye naam pe mat jana aur koi comment bhi Mt Krna plzz this is my mom's id
Waise Jo bolna hoga mujhe bol sakte ho my name is Sagar,
i didn't take this as a quadratic of 5, ill explain it in brief here, i made the quadraric in (1/2 + x ^2) added and subtractited x^2 and 1/4 , other quadratic was made as a result of it, i did a^2-b^2 = a-b)(a+b) and then boom, same two last quadratics as you, this took 3-4 mins of thinking
It should be+10x^2
itna sochne ki zarurat nahi
f(x) = 5-x^2
f(x) = f(inverse)(x)
it implies
f(x)=x
x^2+x-5 = 0
(-1+-sqrt21)/2
one of the easiest approcach and in very less steps is:
root(5-x) = 5- x^2
rearranging x^2 = 5 - root(5-x)
taking root both sides
x = root( 5 - root(5-x))
and we can replace x in RHS by the value of x from the LHS,
so then x = root(5-root(5-root(5-x))) and repeat so on,
and if we look reverse way then,
x= root(5-x), then squaring both sides
then we get x^2 = 5-x, and hence we have solved this SO CALLED AMAZING EQUATION by PATTERN
But this step is a bit risky because you are not sure if the D (b²-4ac) will be a perfect square and if it wouldn't have been a perfect square it would have been more difficult that way.
Search "when mathematician gets bored blackpenredpen"
5 is posive number
Ye achhi community h doston...time agr h 2 saal acche se to follow krna..agr last 3 4 months me ho to follow pyqs only and modules of your coaching..
AIR 1789 (JEE 2022) here
7:23 par quadratic 5 ke terms mein bana lo
Bahut time lag Gaya karib 8-10😢😢😢min
Just take lhe LHS as f(x) and RHS will be f inverse x, they will meet at y=x
Just a 1 min problem😂
☝This is the beautiful solution
I made this in 1/Na seconds
Yahaan (x) ko subject rakh hi nahi rahe hain (hence the title 'value of 5')... Ye demonstration Mr. Aman Malik ne bhi kuch kuch videos mein diya hua hai. Of course, no practice means I'd forget eventually. Lekin memory refresh ho gai 7:20 par!
PS: I didn't take the JEE, Mr.Simplified... I just like math and your channel is really interesting.
Bhaiya graph se 2 solutions hai sahi hai kya 2 solution
Change your channel name to something related to maths.. that way channel will go in a different bigger numbers... Take maths conspiracies and all... Thank me later!
For this can’t we take sqrt 5-x as the inverse function of 5-x^2 and make the eqn be equal at line y=x
Apne under root me (2x²-1)²-4(x⁴+x) liya tha...par aap uss (2x²-1)² ki jaga (2x²+1)² Lena tha ....tab solution pakka hai🔥🔥🔥...well done and good job sir and Aditya too🎉❤❤
5-x negative ni hona chaiye
To, x is less than or equal to 5
Ye ek condition aagyi
Dusri condition h ki 5-x^2 negative ni aana chaiye
Mtlb x^2 is less than or equal to 5
Put krenge to do values reject ho jaengi
Bhaiya yeh toh bohot easy sawal hai ... Trignometry use karke. Take √x=√5costheta maan ke baki toh manipulation hai x nikal jayega ..
graphical method has always been best.
plot both the graphs and since they are inverse of each other, one or more root lies on y=x
5-x^2=y=x (first quadratic)or √(5-x)=y=x
I answered 0️⃣0️⃣0️⃣0️⃣0️⃣0️⃣0️⃣ AT FIRST SIGHT.......... IS THIS CORRECT 💯💯💯💯💯💯?❓❓❓❓
Another method i tried:
subtracting x from both sides
√(5-x)-x=5-x^2-x
rationalizing both sides
[√(5-x)-x][√(5-x)+x]/[√(5-x)+x]=5-x^2-x
simplifying the numerator
[5-x^2-x]/[√(5-x)+x]=5-x^2-x
Cancelling out Numerator of LHS and RHS ( noting that we have a solution where 5-x^2-x =0)[quadratic no.1]
[√(5-x)+x]=1
bring x on rhs and now you can solve the quadratic no.2
Where are other 2 solutions...
That's the only problem about this method...
Be aware next time !!!
WE can eliminate the values based on the domain of x we get . 5-X is under square root thus x must be less than 0 and 5-x^2 must be more than or equal to 0. So we get x must lie between ..... negative root 5 to root 5 . So we are done.
so first i graphed both the functions to check the number of real solutions, they intersected at 2 point (1 +ve & 1 -ve). Now i started solving the sum, i took the x^2 term to the other side (x^2 = 5 - (5-x)^1/2) now sbs we get x = +- ( 5 - (5-x)^1/2)^1/2 now we plug in the value of x in this equation and get a infinite loop of -ve square root of 5 now assume that to be x sbs get a quadratic and we got the positive root now we can use -part of x [-(5-(5-x)^1/2] to get our other solution in a similar way but this time we will get alternate +ve and -ve square root of five in our loop.
√ ( 5 - x) = y = 5 - x^2 , say
5 - y^2 = x = √ ( 5 - y)
Hereby
x + y^2 = 5 = y + x^2
(y - x) ( y + x) = y - x
( y - x) ( y + x + 1) = 0
EITHER
x = y = 5 - x^2
x = ( - 1 + √(21)) /2, - (1 + √ (21)) /2
OR
- ( x + 1) = y = 5 - x^2
x^2 - 3x + 2 x - 6 = 0
x = 3, - 2
Domain satisfie Karo answer mil jaga mara [-√5,√5] aa raha ha wase mana alag method sa Kiya ha answer bhe match ho raha ha
10:19 if we will not then root ke andar -ve aa jya ga?