Calculate the value of 5, yes you heard it right

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Time Stamps
00:00 Introduction
00:43 Approach
05:27 Beautiful solution

[@jeesimplified-subject]

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

[@puchokaun]

Error*

[@yuraje4k348]

@@puchokaun error likhne me bhi error ho gyi bhai se 😂

[@adityaranjan5306]

😅 yeah I have seen this

[@rocketsandmore6505]

@@yuraje4k348 toh kya bhai? hota hai insaan hai wo bhi

[@xgodgaming9515]

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 😢😊😊

[@AayushSrivastava0307]

Copied straight from blackpen red pen 😮

[@ts9dream]

I see a man of culture here

[@IaM_MaNiSh]

Can you give me the link of video

[@adityaagarwal636]

Haan Bhai, vahine se Diya tha maine😊, khud to itne badiya sawal nahi bana sakta na😅

[@UmG_Editz]

​@@adityaagarwal636hello aap ne hi ye sawal bhaiya ko Diya tha

[@adityaagarwal636]

​@@UmG_Editzhaan bhai

[@cblpu5575]

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

[@aarushkumar3972]

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.

[@fmarten02]

I think im only to do like this 😂

[@ryanrahuelvalentine2879]

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.

[@goldensnitch5492]

This doesn’t always work.

[@pkgupta2005]

@@ryanrahuelvalentine2879 oh really

[@JustAnotherSomebody001]

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

[@rupak07ambekar18]

Yeah man

[@tanishpaul0077]

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.....

[@twwilight_gaming3172]

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.

[@MrVarunAgnihotri]

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

[@tanishdungarwal6765]

Lovely sol. bro Congrats on completing the challenge.💯

[@ceezan8085]

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)

[@bhuvneshbhaskar2431]

We have to consider this( -✓5,✓5)domain to

[@miraftabrahaman5628]

Thats correct but must be time consuming

[@VanshamStudy]

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

[@user-bd1ty4uw6r]

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.

[@shauryashikhar3946]

Acha community h bhoi log.... JEE Adv 2016 rank 5574 here❤

[@AbhishekRaj-ji7dn]

Hello bhaiya can u please guide me I just moved to 12

[@iitiandev121]

Aise comment ke nhi hoga ache se guide..bhaiya ka paid mentorship le le @AbhishekRaj

[@AbhishekRaj-ji7dn]

@@iitiandev121 bhaiya wo jyada ho jayega abhi utna afford nhi kr skta

[@reviewer3562]

Konse iit m ho bhaiya

[@AbhishekRaj-ji7dn]

@@reviewer3562 Tanishq rajak?

[@lakshay3745]

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 👍

[@meowman985]

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

[@studyonly4610]

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

[@gamingwithnoob69]

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

[@manjuverma8213]

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

[@kumudsaraswat1398]

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

[@luckygupta4619]

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 .

[@CuriousMango247]

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

[@aditya-1734]

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

[@japjotsingh6327]

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

[@yuraje4k348]

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

[@prince-hb8qk]

Bhai 2 min tak pagal ho raha tha ki x=1 kaam kaise kar raha hai Mujhe laga mai maths bhul gaya

[@KTRYT_]

@@prince-hb8qk same 💀

[@ShivamThanki-qh8wy]

Exactly

[@vivekdubey8354]

Bhai nasha karna band krde

[@technicalstudy6271]

Same

[@Ndanexusofficial]

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

[@DivyanshuYadav05]

Ek doubt hai bhai aapne jo yha per 3:16 x ko liya aise to hum Ramanujan submition ka solution laate hai but wo wrong approach bataya jata hai to yha sahi kaise ho jata hai

[@lakshay3745]

Bhai , baaki videos nhi kroge upload?

[@Gauravyyyadav]

10:19 if we will not then root ke andar -ve aa jya ga?

[@jacob.5849]

hey I'm not planning to give JEE or anything, just a random 9th grader. In the question, since the value of 5 is not equating to 5, since the very start 5 is being treated as a variable right? only in the second solution was an alphabetical value assigned to the variable 5?

[@NicolaTesla28]

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".

[@jeesimplified-subject]

🤔 I thought it adds on humour to but damn, you are right bro Will strongly consider this opinion

[@AbhishekKumar-us6wf]

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.

[@rocketsandmore6505]

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

[@LoveJEEMaths]

Amazing ❤❤

[@ankur-ls2kp]

kay gajab solution h guru dev

[@Prince15141X]

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

[@Sourja]

Awesome method boss ❤

[@aishikadhikary306]

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.

[@dhruvbhatia6808]

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

[@mritunjaysingh956]

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

[@mayankjha6382]

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

[@shivankshrivastav]

Nostalgia of jee days jee adv 2018 1331 here

[@lifeisamarathon2098]

what are u doing now dude

[@shivankshrivastav]

@@lifeisamarathon2098 I'm currently working in AMD as an lithography engineer at TSMC, Taiwan.

[@4fgaming925]

@@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

[@lifeisamarathon2098]

@@shivankshrivastav 🙌🙌hat's off, wishing for your best.

[@aaravmandhanya]

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

[@iitiankavye]

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

[@rationale1734]

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

[@kaivalyajeurkar4604]

7:20 quadratic in 5

[@Izzyw-zw9hf]

App used?

[@criticgamerz6382]

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 ❤

[@manjuverma8213]

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

[@prateeksingh5574]

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

[@HarshKumar-ph8jf]

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

[@HarshKumar-ph8jf]

it was a really nice ques was able to do it with 2 diff methods and got to learn about third one which you told about 8:00 Thankkss

[@pkgaming-wp3pu]

bhai new vids kab aaegi

[@bhuvneshbhaskar2431]

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

[@itachu.]

nice approach

[@ved1081]

how to use you channel if not completed the syllabus

[@shivansh55]

Bhaiya graph se 2 solutions hai sahi hai kya 2 solution

[@ayushkumaryadav2767]

Bhaiya (5-x)^1/2 = (5-x) By taking log yese bhi to kar skte haina

[@Gauravyyyadav]

2 case nhi hoge + or - le ke x ki values nikle gay?

[@phymo4135]

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.

[@adityarathi8503]

2:35 if x²=t then x=±√t

[@RohitGupta_8055]

Itna sundar. Waah.

[@Kuchbhimatlabkuchbhi847]

3:26 , but woh kon saa ek term hai jo kam hoga...??!

[@ryanrahuelvalentine2879]

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

[@kartiksingh10a24]

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

[@jeesimplified-subject]

Glad to know bro 😁

[@user-ub6ok1yr4u]

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

[@b4bhavya770]

I did with same method

[@bhoju_]

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

[@abhigyanshree5676]

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

[@rajsah7755]

It is in Allen module function 12 question about no. Of sol

[@jagdishmohapatra4744]

Wow😮

[@adityarajsharma6029]

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

[@VRZ1105]

7:09 Assuming a quadratic in 5

[@sleha4106]

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

[@sahilnarkhede2943]

Bhai please ek series coordinate geometry pe banado advance level problems ki i am loving your series

[@PutushKumar-xq1ch]

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

[@VinyaAhuja]

great qn !

[@cubingdom3426]

9:02 mei perfect square kaise bana?

[@Schrodinger0]

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.

[@iitiandev121]

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

[@SnehalIngale-uc2dv]

👍

[@vihaanbansal2155]

Bhaiya , in the solution doesn’t that equation mean that 5 is a root of the quadratic equation and 5 will take value either x^2 + x or x^2 - x + 1

[@rocketsandmore6505]

badiya question

[@tanmaygarg6869]

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

[@Prateekch4529]

Pw ashish sir ne same approach se ek question class me karwaya tha jisme constant ko variable aur variable ko contant mana tha

[@what_the_fox_officiall]

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..😁

[@jeesimplified-subject]

Good thought

[@shoryagoel1411]

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

[@noexistence88]

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

[@noexistence88]

its plancks constant

[@missionaryrav628]

Woh kiya hota hai?

[@bhoju_]

😂😂😂🤣​@@noexistence88

[@ramanraghuwanshi47]

Plancks constant h ~ 10^-34 1/h~ 10^34 10^34 seconds means 10^26 years !! Think once....

[@abhirupkundu2778]

bhaisab, tum J^-1s^-1 mei banate ho question? Kya logic hai.

[@anuragsingh9767]

Basically

[@rudresh4238]

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

[@pushkarnarware]

😱🤯 pehli baar dekha is type ka kuch

[@MitulPrajapati-el6sb]

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

[@user-hj7zo2kt5y]

It should be+10x^2

[@___shahid___1950]

3 degree ko ku nhi kr sakte?

[@aaminarahman2308]

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

[@AnasArfeen]

fir, ab solve karte baith polynomial

[@itsrahul644]

Bahaiya main jab video pause karke solve kar raha tha main vi a wala maethod laga raha tha let 5= t But mainne sridhararcharya formula laga diye 😊or equation complex ban gyi...uske bad solution dekha

[@user-dd8zp7vf6k]

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.

[@jeesimplified-subject]

oo bhai 😦

[@ParthBnsl-iitis]

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]

Nice approach bro... But can we not solve by putting x=5cosΩ Ω€[-π/2,π/2] ... My process is short

[@ParthBnsl-iitis]

@@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

[@unbeatableayush3766]

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

[@Mathlover_1729]

@@ParthBnsl-iitis yes bro,this is the problem

[@BigMan-il8dj]

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.

[@sangeetatiwari5088]

Lekin when we put x=root 5 it create contadiction

[@xgodgaming9515]

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 😢😊😊

[@ClarkKent-bz9tf]

ex function hai dusra uska inverse they dono x=y pr intersect krenge

[@bharadwajkk6823]

AS bprp said, believe in the power of geometry, considering making a triangle with said thingy

[@gangster609]

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