Very Smart Solution Of JEE Mains 2020 Algebra Question | Bhannat Maths

[@iamliterallyvengeance]

just a trick I wanted to share I think people feel easier to solve quadratic eqns having coefficient of x² as 1 for example take 9x²-18x+5=0 so the trick is just multiply coefficient of x² to constant term by and make coefficient of x² as 1 now the equation is x²-18x+45, solve this ,and then just divide the roots of the new equation by the original coefficient of x² and there are the roots of the original equation

[@28aminoacids]

Thanks for the trick. Never thought like this. I'll definitely use this in my upcoming exams.

[@santoshichoubey4075]

Bruh aise bhi toh 45 ka dhundhta na Bhai abhi bhi wahi hoga na

[@iamliterallyvengeance]

@@santoshichoubey4075 nahi samjha tum kya kehna chahte ho?

[@fearlessplayz2800]

Mod x ke liye hi use kar sakt hai kya sirf bro

[@iamliterallyvengeance]

@@fearlessplayz2800 kisi bhi quadratic equation me use kar skte hai kyunki after all |x| to t put kar skte ho

[@parameshwarhazra2725]

I am graduate student still learning so important things from you Aman Sir. Love from बंगाल। ❤🎉❤

[@honestadministrator]

(18 |x| ) ^2 = ( 9x^2 -5) ^2 or 81 x^4 - (90 +324) x^2 + 25 = 0 Hereby products of all roots is 25/81

[@pwhike]

SIR AAPKE VIDEO DEKH DEKH KAR AAJ IIT ROORKE CSE PAHOCH GYA THANKS LOTS DEAR ❤❤❤

[@Hustler_Explorer]

Bring more such videos frequently sir❤

[@Shivam_7727]

Concept dekh ke maja aa gya... brilliant question for brilliant student👌👌

[@SK_LynIx.7k9546]

I love such videos of yours, they are such a great help for our preparation Please make more such videos

[@visheshtiwari7555]

What a concept

[@atharv8708]

It was a very easy question of basic maths

[@MalcomX852]

Ashish Agarwal sir has already done this question in last year's Arjuna JEE 2.0 in the basic mathematics section....Arjunas can solve it without pen and paper🤷‍♀🏌‍♂.

[@Theaimers07]

True tarun khandelwal sir ne bhi krwaya tha basic maths me without pen paper solve ho gya ye

[@ashutoshpani6944]

*Sir please make a video on Sigma or summation* we are facing problem because our concept is not clear ....... Pace series 😁😊😊 me ap banaye the samajh me ayatha but ap questions jiyada nahi karaye the ...... If you have time sir please make 😇😇 Now a days I'm missing your lectures in bhannat maths ..... Aj bhi pace series mere dil me basi hui he sir 🥺🥺. Mere piyare Aman sir kuch sooch ke chore honge pw 🥺. .... Fir bhi ek bat Jan lijiye ap maths ke sher ap hamesha rahenge ...... Bhagwaan kare mere hise ki sari kusi apko dede 😊.....

[@shivanshkapur4872]

X>0 9x²-18x+5=0 X=18±√(324-180)/18 =18±12/18 X=3±2/3 1/3,5/3 X

[@gigageta3273]

Same method tha Mera 🗿🍷

[@PS-mh8ts]

one can use the identity x²=|x|² to solve like so: 9x²-8|x|+5=0 i.e., 9|x|²-8|x|+5=0 let |x|=u we get: 9u²-8u+5=0 => (3u-1)(3u-5)=0 i.e., u=1/3 or u=5/3 but u=|x| i.e., |x|=1/3 and |x|=5/3 both these values 1/3 and 5/3 are +ve, and hence possible values for |x|. i.e., both 1/3 and 5/3 are valid solutions. thus, we get: x=±1/3 ; x=±5/3 product=(1/3)(-1/3)(5/3)(-5/3)=5/81

[@shivanshkapur4872]

@@PS-mh8ts tab bhai itna idea bhinahi tha and vaiseh i prefer cases analysis rather than overalls analysis

[@vnpskarvindmule7246]

Acha video tha sir main chahunga ki aap aise video banate rahiye...........❤❤❤❤

[@Theaimers07]

Very easy question aur ye question basic maths me mod seekhne ke bdd easoly solve ho gya dekhkr hi kr diya mind me Confidence mila krke❤

[@IssacNewton-be2ik]

Really jee make good question I really like to solve these questions

[@vikashkumar96314]

Sir Quadratic equation ke aise hi kuchh aur tricky questions ko lekar aaiye sir Plz ye ek regular customer ka request hai har video dekhta huu aapki

[@reetadevi6262]

Sir how did [x1x2] equalises to [5/9]² please explain this

[@darkeyes9902]

as x^2 >= 0 for all real x, then |x^2| = x^2, replace x^2 with |x^2| and use the fact that |x^2|=|x|^2 and you get quadratic equation in |x|, solve & find roots then multiply them.

[@bhaktiwithpayal2913]

I just want to share my approach We can take two cases first where X is greater than zero when modulus breaks positive and when X is smaller than zero where modulus breaks negative When modulus breaks positive the quadratic which is formed as two roots both of which are positive from observation therefore both the roots are valid and we take the multiplication of both the roots when modulus breaks negative we get another quadratic equation and both the roots of the quadratic are also negative so both of those roots are also valid so it take that multiplication product of both of those roots are 5/9 so 5/9 squared 25 by 81

[@immanbeyondlimits.....4825]

Sir mai Daave ke sath kah sakta hu ki agar jise maths se prem nhi to aapke lectures dekhne ke bad ho jaega .... Sir abhi 10th Board karke aapke purane pw ke lectures dekhe Kya bhannat padhate ho aap Functions relation pura clear ho gya .... Thanku sooooooo soooooo much sir ❤❤❤❤❤❤❤❤❤😊😊😊😊😊😊 😁😁😁😁😁😁 Bahut badiya padhate ho aap Upar wala aapko hamesha khush rakhe ..... Ab 11th pura aapke videos se hi karunga ❤❤❤❤❤❤

[@Satyamraj522]

Kon kon sir ko pace series me miss krta hai 🥺

[@visheshtiwari7555]

Me bro

[@anshswaroop6849]

Nostalgia 🥺🥺🥺

[@ritishprajapati9205]

Mai bhi❤

[@VijayKumar-be5it]

Hu.....

[@AlokKumarSingh-fy4uz]

Me

[@YashChaudhary-b8q]

Sir ajj he Arjuna batch me ye question kar aye gya hai

[@Dhanika112]

Donesir Concept used |x|^2 =x^2 solved it by substitution and got the solution But sir your method is great ❤

[@aryankushwaha7028]

Sir there is one question x²+x-2x and the zeros are m ,n ans we have to find (1/m )-(1/n) , and its answer is ±3/2 but in many books answer is only -3/2 why sir please tell sir

[@parv1706jain]

Sir meine ye kal subah he pyqs mein solve Kiya tha aur Aaj aapka bhi video aa gya.

[@AdityaAhire-f8w]

These are literally free marks NTA is handing out to students, so students shouldn't miss such simple questions which can potentially increase their rank.

[@ojasviagrawal8109]

😂 aise toh poori physics free marks h

[@AdityaAhire-f8w]

@@ojasviagrawal8109 Sahi kaha🤣🤣

[@jaseelvp]

​@@ojasviagrawal8109 Nowadays 😂

[@penspinner7926]

@@AdityaAhire-f8w konsi class hai apka?

[@Jeejunction80]

Sir aap to gajab hai pls aisi aur videos laaie

[@hitenrajsingh3041]

Sir advance ke acche concept ke pyq karayie.

[@diya2537]

sir apki ye sayariya karna muhje personally bohot or motivating lagta hai lekin apka vo video suru karne se phele vala dialogue ki muhje yad aati hn !( toh kese hn log umid hn ache hoge or me bhi ho bhaanaat! ) sir apke channel ka nam bhanaat hai lekin ap apni purani phehechan bhol gaye ho!

[@gauravdhakate25]

Please tell which digital board you are using?

[@AnjaliKumari-lu6xf]

Yes sir i want more videos like this

[@mdemrankhan2858]

x^-1 = 1/x , why? Please sir, make a video on this topic.

[@navnathlonkar2240]

Study law of indices bro

[@studyacount-uv6uj]

​@@navnathlonkar2240 he is probably not asking the concept but why mathematics work this way?.......its like asking why -ve×-ve is +ve

[@Rudra-Raut811]

@@navnathlonkar2240 he is maybe in 10th cbse who has just started grasping maths

[@satyamsrivastav1772]

Sum of roots = -b/a. Product of roots=c/a.

[@Mayank_Guglani91]

Thanku so much sir aise videos roj banaya kro Kam se kam 2❣️

[@thoughtsofs7728]

Sir kya ITF equation is part of IIT

[@surajdey2877]

Make more videos...i really enjoy your videos

[@ajdjdhdhshahmajaj]

Sir I am from Pakistan and I have seen your most videos

[@Someone_21]

Atta mila abdul 😂😂

[@Someone_21]

@@umbrella9801 Yehi chutiye log h jinki sarkar terror ko fund karti h. Yehi chutiye ke tax ke paise se woh fund karte h terrorists ko. Poonch par hamare 5 jawan mare in chutiyo ke karan aur tum jaise log inhe eid Mubarak aur love from india keh rahe ho 😒 Yaha kaha par political party aa gyi, army pure desh ki h kisi party ki nhi. Inhe chutiyon ke paise se hamare 5 jawan mare h poonch me

[@Adityathakur.-_-_-_....984]

Video tone op sir❤

[@KUNDANKUMAR-yw1hw]

Sir apne app per jee ka full batch kb lauch kijiyega

[@omprakashbhadoria1157]

Very good problem, put up more

[@shutosh2006]

Thanks for this new concept to me

[@ashokkhullar6650]

Another method may by taking 18x to R.H.S. and squaring both sides to get the equation 81x^4 - 234x^2+25=0

[@PS-mh8ts]

squaring a given equation always poses the danger of introducing extra solutions. for example, if we square both sides of x=5, we'll get x²=25, which has 2 solutions, whereas the original equation had just one solution. the most straightforward method (and hence fool-proof and requires no guess-work) of solving the given equation is to use the identity x²=|x|² and solve for |x|. thus, we have 9|x|²-8|x|+5=0, which is a quadratic in |x|. solving it, we get |x|=1/3 or |x|=5/3. both values are positive and hence legal values for |x|. thus we get: x=±1/3 ; x=±5/3... and the product is thus=(1/3)(-1/3)(5/3)(-5/3)=25/81

[@IITJEE2024-m7r]

Yes sir as a such video to help much

[@soumikchatterjee9345]

Let , |x| = t or, √(x²) = t Or, x² = t² Now , 9x² - 18|x| + 5 = 0 Or , 9t² - 18 t + 5 = 0 Or , 9t² - 15t - 3t + 5 = 0 Or, 3t ( t-5) -1( t -5) = 0 Or , (t -5)(3t -5) = 0 Or , t = 5 & t = 5/3 Now , the product is 25/3

[@maroofrather7633]

Quadratic achaay se solve kr le Bhai galat solve Kiya ha.

[@prashantabasak4772]

Sir, kya hum iss question ko aek aur tarike se analyse kar sakte hai like : iss wale quadratic equation mee 'c' ka sign positive hain and 'b' negative hain. Also product of 'a' and 'c' is positive. So, hum yeh bol sakte ki jo middle term hai jo ki negative hai usko hume sum of negative numbers hi likhna parega(by seeing the sign of 'c'). Hence iska matlab yeh hain ki mod 'x' positive hain and then hum further proceed kar sakte hain... Plz reply sir. Puskar Basak.

[@amishmittal4536]

incorrect hai bhai

[@prashantabasak4772]

@@amishmittal4536 Can you plz explain ki kaha galti hai??

[@nityajangra]

Really saying I did it in almost 7 to 8 sec !

[@mordernphysicsspshukla1773]

sir plz make on video on higher algebra arihant bernald child book 🥶🥵👍👍

[@mrindian9316]

Sir you are great teacher 💕💕💕💕💕

[@doraemonnew1610]

Here 9x²+5=18|x| Square both sides.Mod gets removed. Simply apply products of roots d/a for the biquadratic equation.

[@shhakks]

Great

[@adarshandayushplayz7436]

Great approach

[@digantpanara08]

Apne pw chhod diya kya? Plss batao sir

[@GaneshSharma-fv3nb]

Yes sir ji or video ❤

[@yashkukreja3049]

2020 ka Kyaa samay tha badiya asaan se sawaal aate the quadratic equation se ab pata nhiii saala 2022 kya hogaya haii naam sirf quadratic haii sawaal toh majorly biquadratic se puchte hai aur meri shift m April attempt m toh had hi paar kar di X raised power 8 k roots par sawaal aa gaya 1 usko solve karne m saala 5 baat shridharachaarya lag gaya 😂😂😂😂. Hey bhagwan kyaa ho raha haiii teri iss sundar si bnai duniya m 🤣🤣

[@abhinandandhakad4169]

Sir please bring more sessions ❤

[@nikhilraj9544]

Advance problem ka solution upload kar deya sir please jee mathematics black book

[@akashthebot7265]

I got it the right ans.lov you sir🖤🖤

[@ujjwalghosh389]

Sir please bsc maths ki bhi class ap start karyaye na 🥺🥺🙏🏻🙏🏻

[@rajajiii_001]

Sir nda ke liye batch nikaliye na please

[@avipanchal8580]

Sir esi or videos chaiye

[@honestadministrator]

Product of the roots remain invariant irrespective of +ve x or -ve x And equals to 5/9

[@comptech5240]

that is the product of just 2 roots. You are getting 4 roots of the equation.

[@kumaramit04-q6c]

Yes, four roots because of 18|x|. b can be positive of negative

[@rishabhyadav9986]

Sir |x| to +ve x ke liye = x deta hai or - ve x ke liye -x deta hai to aap kese bol rahe hai ki |x| ki value - ve ajati hai to usse reject kar dete hai 3:21

[@sumit_625]

Sir This is Bannat sir ka bhannat solution

[@meraki8702]

25/81

[@ajaybishnoi4967]

Sir please solve e to the power x square = i .

[@saurabhsuman1939]

Yes sir chahiye

[@navamgarg]

Mazza aagya sir 😳

[@ashishkumaryadav8735]

Ye kis class ka concept hai

[@sankarsanyal3118]

log series e series bhi karao

[@pulkitarora28]

Awesome sir❤❤

[@raghavgoyal1914]

Sir Careerwill me jo Determinants padhaya hai wese hi Sare chapters padha do 🙏🏻.. Batch launch kar do 5-6k rupiye tak ka please sir 🙏🏻

[@adityagamer8012]

Yes sir i wants these😊😊 types of problems ❤❤❤❤❤

[@sawanverma-b8r]

sir you are the best

[@mathsloverprashant9109]

Violet Vape 🎶🗿

[@TejasIngale-vu3ny]

Very easy question ❤❤

[@mdnasirazam8371]

Sir ISI and cmi exam k liye bhi humari help kar dijiye kuch hi din rah Gaye hai exam k liye

[@sukritineogi8195]

Yes sir

[@ThemeloD-so5wd]

Salute

[@ANKUSHKUMAR-gr8tp]

Yes sir please jee main and advanced 2025 ka batch laiyena sir please 🎉🎉🎉🎉🎉🎉🎉🎉🎉

[@divyanshdubey0740]

Love you sir

[@ArshSahu]

It took me 10s to solve the question just by looking at the thumbnail. I didn't even see the options and I'm in 11th rn

[@LuckyPhodaga]

Mujha to option bhi yaad tha 😂 yeah I am too in 11th

[@sukritshankar03]

[ The Key is not the modulus here, but that the function is even ] Since function is even, it has symmetry about the y-axis, meaning roots r1, r2 in the case x > 0, will give roots -r1, -r2 in the case x < 0 !! Multiplying all roots will give (r1r2)^2 and the product of roots can be seen from any case to be 5 / 9, and then we arrive at the answer by squaring !! This is possible because the function is even (f(x) = f(-x)), but the function is even not because of the modulus, but due to the overall composition !! This point, in my view, should be clarified and emphasized, else the explanation here is honestly misleading !!

[@tanishk_bajpai__]

Sir please √dx ko solve kr dijiye🙂

[@piyushmortal8161]

Nice sir

[@dhritishmankumar881]

Aman sir ❤❤

[@sandeepgamer9491]

25/81 ?

[@rajumaths_]

Mja aa gya question me

[@nandakumarcheiro]

sum -b/a product c/a ans 5/9.

[@impresent2005]

💕💕💕

[@sahilbatt]

Sir koi Instagram par aapki picture use kar ke JEE ka course sell kar raha hai Instagram ads kya through Rs. 60000 ka. Maine us ad ko report Kiya hai.

[@papadeltaa-rc3jf]

Sir aap courses kyu nahi launch karte aapne pw chod ke bahut badi galti ki

[@Communalसनातनी]

as a 10standart student i choose b

[@Ashfiqueahmedbrohi-hf9jj]

Sir product of roots ki to -b÷a formula ho ta ha ais formula as to answer 2 ayia ha

[@shhakks]

Chalbe lodu aata h nhi gyan chod rha h

[@Rudra-Raut811]

are bhai me bhi 10 me hu apko yaha sikhaya hoga ki products of roots c/a hota he par isme mod ki vajha se unko square karna padta he jie ans 25/81 ata he mene foundation kar raha hu isliye pata he

[@arunsingh-ny7uh]

Sir bahut mast Content laate ho pura feel aajata hai maths ka❤️❤️ jab ki muje maths bilkul nhi aati 😓

[@YaldaShankar]

[ Arre Aisi Ad-hoc baatein mat sikhao ] The straightaway concept is that for x < 0, we have an equation, and x > 0, we have an equation - Find roots in both cases, and multiply all !! Now, it's easy to see that the function is even since f(x) = f(-x), so there is symmetry about the y-axis !! Thus, roots m,n in x > 0 case, will become -m, -n in x < 0 case, and then product becomes (mn)^2 = (5/9)^2 (Take product as 5/9 from any one case) !! You still get the answer without any calculation !! But you get to (mn)^2 because the function is even, not because of | x | !! Just imagine JEE tomorrow changes this to 9x^3 - 18 |x| + 5 = 0 (A very Simple Change) !! Then your whole theory here goes to sky, and people will suddenly be confused. No longer you can say that the solutions will be |x1|, |x2|, |x3|, since the function now is not even !! In that case, you solve equations for the two cases and multiply the roots !! Even functions have y-axis symmetry, and odd functions have point symmetry around the origin, which does not necessarily translate to roots !! There are functions which are neither odd nor even, and root symmetry can't be guaranteed there !!

[@hitlarjha5708]

Please sir esi type ka video bhej yeh kuch naya shikne ko milega please

[@manormachoudhary7096]

Wah hitler wah kya baas kahi hai

[@lalitarajoriya6628]

I have also a trick: as sir has said that the equation must have four roots, then the only number factorizable in four factors is 3rd option so its the answer😅😅😅😅 lets see what you guys say

[@coding427]

I like this video .

[@alokthakur9434]

💗💗💗

[@bruh_its_me]

Bhannat!

[@ajal3025]

Sir mod x ko t let krke kiya within 30 sec ho gya

[@ajal3025]

@@mohammadalfaraz1027 hn

[@gigageta3273]

2nd attempt ma solve Kiya ha is question maina 🙂