Calculate the value of 5, yes you heard it right  

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

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

Reminds me of that problem in "pair of straight lines", applying the concept of homogenization of a curve, taking 1 as a variable.

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.

Copied straight from blackpen red pen 😮

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)

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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.....

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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

7:04 Understood, we have make a quadratic taking 5 as a variable and finding it's value in terms of x.....❤

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

10:41 cases orignal eq se eliminate ho jayenge since x² is less than 5 therefore x would be approx less than 2.23🙂

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.

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 👍

Are gajab Eise sochna bhi ho skaata Q chota sa hn par Eise karke kai equation mein apply kar sakenge Thanks Bhaiya ji

1:28 apply rational root theorem agar sare zeros irrational and complex nhi h to work krega

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)

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

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 = 5-x² Squaring both sides 5-x=25+x⁴-10x²

X = 1 kaam kaise kr rha hai ? Check kro equation me x = 1 daalke

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

X=1 toh root hai hee nahi?Factor kese kar diya uska

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

haaa bhaiya main hi hi 1/ plancks constant sec me banane wala

Lovely sol. bro Congrats on completing the challenge.💯

another easy and quick method: let under root 5-x be y hence y squared = 5-x so x=5 - y squared. let this be equation 1 5 - x square is also y. y= 5 - x square. let this be equation 2 so equation 1- equation 2 will be x square - y square = x - y so either x=y or x=-y putting y=x in equation 1, we get x=5-x square. this is quadratic and can be solved putting y=-x in equation 2, we get y square= 5+ y. we can solve for y and hence get x becuase x=-y

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.

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 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 .

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

Since 5-x²=(5-x)½ we have 5-x²>=0 ie -(5)½

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

Here’s a way to solve the initial equation obtained of 4th degree: X^4 - 10X^2 + X + 20 = 0 Factorising the 4th degree expression into two quadratics: Let, X^4 - 10X^2 + X + 20 = (X^2 + AX + B)*(X^2 + CX + D) Comparing coefficients of X^3 in LHS and RHS, C+ A = 0 or C = -A Comparing coefficients of X^2 , D+ AC + B = -10 and because C = -A, hence D- A^2+ B = -10 -(1) Comparing coefficients of X, AD + BC = 1 and because C = -A, AD - AB = 1 or A(D-B)=1 -(2) Lastly comparing constant terms, BD= 20 - (3) Trying for integral solutions to (3), we obtain B = -4 and D = -5 (by checking divisors of 20) Putting B = -4 and D = -5 in (2), We get A = -1 Now it remains to check if these values satisfy (1) LHS = D- A^2+ B = [-5] - [1] + [-4] = -10 = RHS. Therefore, A = -1 B = -4 D = -5 And C = -A = 1 Hence, the original 4th degree expression can be factorised as X^4 - 10X^2 + X + 20 = (X^2 - X - 4)(X^2 + X - 5) Now to find the roots, (X^2 - X - 4)(X^2 + X - 5) = 0 The roots obtained are [1 (+-) sqrt(17)]/ 2 and [-1 (+-) sqrt(21)]/ 2 As the domain of the function can be found to be [-sqrt(5), +sqrt(5)] The acceptable values of roots are 1/2*[1-sqrt(17)] and 1/2*[-1 + sqrt (21)] QED. This is not a general method, but often works for depressed quartics.

7:20 quadratic in 5

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👍

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".

7:09 Assuming a quadratic in 5

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

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 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.

X= (-1±√21) /2, given expression equates a function with its inverse, and we know that a function and it's inverse always meet at line y=x, therefore we can equate 5-x^2 with x to obtain the ans, but we have to check that whether they satisfy the condition and on checking we will obtain our final ans as X=(-1+√21) /2

another method we can do is creating prefect squares ....shift root(5-x) to the right add and subtract x we will see formation of 2 perfect squares {root(5-x)^2 - 1/2}^2 = (x- 1/2)^2 ...now we can easily solve these equations as it will reduce into 2 quadratic equations..😁

drawing graphs of both the function simultaneously can give no of solution. For value your method is great

okay, i loved it, i watch a lot of bprp and mind your decisions, post more such ques, loving these

It should be+10x^2

7:30 made two perfect squares which came out to be of form a^2-b^2 than factorized it to two 2nd degree equations and hence got the 4 values of x

2:00 cardan method Ferrari method Kya kre fir😂😂😂

I have solved such question on bhannat maths channel in which he habe given cubic so it was eazy for me

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 😢😊😊

eliminate kiye using 1st equation 5-x^2>=0

One doubt. Agar quadratic 5 me banali humne then sum of roots and product of roots kiske equal hoga?🤔🤨

kay gajab solution h guru dev

2:26 if x^2 = t then x = +- root t btw thanks for the question

Jo keh rhe h unse 1 baar me solve ho gya khud se Meanwhile their jee main %ile

(1 - √17)/2 and (√21 - 1)/2

Yes 7:10 5 mei quadratic man kar x ki value nikal na😊

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

7:27 solve by dharacharya and 5 as a root

7:28,understood,we need to take 5 as a variable and assume it as a quadratic

People laugh at me whenever I have tried some similar approaches like this😢😢 They think I'm just crazy.

Bhaiya, y=√5-x and y=5-x² inverse functions hai.Then y=x pe unka root lie karega.

5 ko variable ki tarah treat kiya aur x aur uske saare terms ko contsants bana diya. Aur uske solutions use karke x ki values bata di. x = (-1±√(21))/2 or x=(1±√(17))/2

Its almost wondeful to see people solve such beautiful equations, i saw this same equation 5 years ago on a channel black pen red pen. (Dear tejas i solved it or something like that was the title). Amazing to see this questions again after so long.

I subtract -x both side and take (5-x )as a variable and then solve it and then it become just like a normal linear inequality

Bhaiya yeh toh bohot easy sawal hai ... Trignometry use karke. Take √x=√5costheta maan ke baki toh manipulation hai x nikal jayega ..

Bro when pointed out that it’s quadratic of five then it ringed very well thought loved it may be I can get types of question related to this that will be fun . Anyhow I enjoyed the solution very much it’s not everyday u get some nice ideas when solving

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

7:23 par quadratic 5 ke terms mein bana lo

anyone who has watched the video how real men solve equations knew this approach and it was discussed by blackpenredpen as well

Quadratic in 5 banti lag rahi hai.Title helped me

√ ( 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

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🎉❤❤

We can also convert this to infinite nested root and we can solve that easily

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 😊

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.

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 ! I did thus que by geetting two parabolas and plotting them under the domain -✓5 to ✓5 (i saw one more guy with thus approach but he may get more roots is not considering the x domain) x°2-y^2=x-y so[ x-y =0] or [x+y=1] Thanks

5 mai jab quad banaya ,,,tab toh dikh gya ,,, but pehle dikh pana was damn tough ,,,but koi nhi ,, now i know a new approach

X=(1+√17)/2 lene pe original equation me put krne pe RHS -ve arha h jo ki nhi hoskta because under root of positive known quantity is always positive so aise ek case eliminate horha h , baki 2nd nhi click kra

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.

Hi bhaiya I am also a jee aspirant aur maine aapka channel shorts pe bahut dekha h aur thank you for giving itne sahi question kuch kuch toh easy lgte h but kuch mein aisi band bajti h but thank you Also main ek request kr rha hu ki please agr ho ske toh aap vaapis vo problem solving ki videos of every chapter lao na 😊 vo bahut acchi series thi please agr ho ske toh please laiye

Bhai , baaki videos nhi kroge upload?

fx=f'x give seen easily it implies that both of them equal to x find x easily by quadratic formula

For elimination any value thats between negative root 5 to positive root 5

Simply range nikal sakte hai equation ki which is negative root five to positive root five

5-x^2 has to be positive so roots (1+rt17)/2 and (-1-rt21)/2 are rejected

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

x-5>0 in the second case thus, x>5

Dhakkan ho kya? Shuru ke 2 mins main hi itni badi galti.... x = 1 kaise solution hai iska? Matlab kuch bhu......

2:35 if x²=t then x=±√t

8:50 me (2x²-1)² kyu hai (2x²+1)² hoga na due to b ka value

Bhaiya koi aur series bhi start kariye please 🙏🏻🙏🏻🙏🏻🙏🏻

5 is posive number

maine ye socha ki pehle squaring karle and then quadratic jo ban rhi usko solve alag kare and =x^4 kar de then x^4 ko x^2x^2 karke jo factors bne the qudaratic ke usse equate akre to same a jayega at 9:24

Bhaiya not gonna lie , at 7:08 when you wrote that equation , it stiked me to complete the square and everything followed up , btw amazing one ❤

It's a inversely fn concept 5-x^2=x and solvr

Y=under root 5-x leke 5 ki value nikalke substitute Kar sakte hain original eqn mai cus usne fir x-y will cancel out after factoring

7:24 PE SMAJH AAYA AND MAZZA AA GAYA