I love these solvers to check my work or to do tedious calculations. Any others I should know about? The best part is they keep getting better. In October I made a video A Problem WolframAlpha Can't Solve, But You Can (615 + x^2 = 2^y) : kzread.info/dash/bejne/doN9tcx_pszXm7A.html. Amazingly a month later someone said that WolframAlpha now does give the correct answer--so perhaps my video alerted them to a bug, or they decided to update the solver. Either way, it's amazing the kind of math technology we have access to today, and the tools just keep getting better.

@rachitrastogi5399

5 жыл бұрын

Happy New year

@vinceguemat3751

5 жыл бұрын

There is a big mistake : 0^0 HAVE A VALUE, it’s 1 ! 0^0 is only an undermined form only for LIMITS so while this question is not about limits we don’t need to check if y != 0 in case 2

@some_one_appear_and_said

5 жыл бұрын

Wow, luckily i have just learnt that kind of exam for a few days. So, easy peasy lemon squizzy :))

@Sam_on_YouTube

5 жыл бұрын

@@vinceguemat3751 Setting aside the validity of 0^0=1, there are no 0^0 scenarios involving integers.

@harshranjan8526

5 жыл бұрын

My mathematics teacher had taught me the same, and almost in the same way also, this question is really amazing.

"You might not believe it, but the answer is 7!" what a nice way to put factorials and trick everyone lmao

@barneystinson3358

5 жыл бұрын

How's that work?

@rarelyseen2609

5 жыл бұрын

@@barneystinson3358 yeah, i dont get it

@herrebner1772

5 жыл бұрын

@@rarelyseen2609 the solutions are 2 3 4 5 6 7, so its 7! (7 factorial), so just a little wordplay ;)

@heitoho5430

5 жыл бұрын

omg

@heitoho5430

5 жыл бұрын

I just spotted it

"A math problem most computer can't solve, but you can" Me: No, I don't think I can

@3characterhandlerequired

4 жыл бұрын

My thoughts exactly. It has been about 35 years since I calculated anything like this but I remember that working with math was fun at school. However in this video there is one jump I don't know how you got what you got and that is how you get (x-2)(x-5)=0 from x^2-7x+11=1? Can someone explain that to me? Sounds like that is something you can do just by looking at the math so it probably is simple thing that I have just forgotten, but I can't figure that out. So "No, I don't think I can" is quite real for me.

@benhur2806

4 жыл бұрын

@@3characterhandlerequired Subtract 1 from both sides, for x^2-7x+10=0, which you can then factor as (x-2)(x-5)=0

@3characterhandlerequired

4 жыл бұрын

@@benhur2806 Thanks for the answer but that doesn't really give me solution to my problem. It's the factoring part that I'm missing. Any websites where I can refresh my math skills (pretty basic stuff I realize now, I have forgotten more than I was willing to admit). I feel like the time I first encountered Laplace transforms. It was in physics class, not math and that was the first time I or my classmates had seen the thing. "You put L in front of that and the whole equation mysteriously changed? What's going on?"

@Alians0108

4 жыл бұрын

@@3characterhandlerequired Try Khan Academy or MathTutorDVD

@benallen464

4 жыл бұрын

@@3characterhandlerequired look up the quadratic formula, it gives the root of any x^2 polynomial (numbers that if inserted into the polynomial make the equation equal to 0)

The 1st and 2nd solutions easy but damn, I completely passed over that 3rd case

@volodyanarchist

3 жыл бұрын

I am the only person who has also considered i^(4*k) and (-i)^(4*k), but completely missed that raising to 0th power gives you 1.

@brendapaolacantubeltran3919

3 жыл бұрын

I actually thought about that but for some reason I got 13 instead of 12 and got 5 and 8 as solutions instead of 3 and 4

@edwardmao9792

3 жыл бұрын

same

@deedeeen

3 жыл бұрын

@VolodyA! V Anarhist "Solve for all *real* numbers x". ;p

@volodyanarchist

3 жыл бұрын

@@deedeeen That didn't stop me. I cannot remeber why.

Not gonna forget about case 3 for the rest of my life

@riem9740

3 жыл бұрын

LOOL

@koleso1v

3 жыл бұрын

Because it is a wrong solution :)

@koleso1v

3 жыл бұрын

@Zoltan Stromberg the explanation is simple. There is a difference when you solve the equation in integers or in real numbers. For integers, all solutions are valid, because indeed, (-1)^2n=1. It is not the case for real numbers though, because when you write a^b, where both a and b are real, then you immediately assume a>=0. Indeed, if a

@koleso1v

3 жыл бұрын

@Zoltan Stromberg but the problem is not in two possible values, it is much deeper. If we consider a power function a^b, we would like to operate with b, as we always do with numbers. For example, (-8)^(1/3)=(-8)^(2/6)=(-8)^(3/9), etc. However, there three values of (-8)^(1/3) = {-2; 1+i√3; 1-i√3} and 6 values for (-8)^(2/6) = {+/-2; +/-(1+i√3); +/-(1-i√3)}. Now you can see what happens. Even for rational powers we can generate as many solutions as we want. In complex analysis, this is generalized to a^b = exp(b*ln|a|*b*(2πin+i*arctan(theta)), where n takes all possible integers and yields to infinite number of solutions. Thus, we have only two regular approaches. 1. Postulate that we want to consider only real solutions, therefore a>=0 forever, and negative a are not ever considered. 2. Properly extend the expression to the complex plane and show all infinite solutions. In this video it was done something in between that is not rigorous and not right, because the (-1)^2k solution is from the different realm, and all other solutions are magically ignored.

@TheMartian11

3 жыл бұрын

wow, I aspire to learn all this stuff one day, someday haha. I can understand the ambiguity that you stated here, but I lack the knowledge to understand anything else as whatever I know about complex numbers for now, is that its "a hack for computing negative roots"- as my maths teacher explained it. I know that's not just all, and that it's a whole new field of maths, but that's just the educational system here, incompetent teachers who didn't achieve their desired life goals (and thus reluctantly became teachers) who don't care about the student's intellectual understanding. they just care that we can solve the textbook questions and use so-called "tricks" to solve MCQs. I'm 15 by the way. :)

0:54 - "well, imagine you are stuck in this problem" Ok, that part I got right

@normaaliihminen722

4 жыл бұрын

Vinícius de A Batista I have watched this video 3 Times And stil I don't* understand what and what is Point of this. I really want to understand Math but i have learning disability called f80.

@glumbortango7182

4 жыл бұрын

*these math videos inspire and build confidence for people around the world*

@eelueelu9069

3 жыл бұрын

I did this question in first 15 seconds..believe me I got the roots of these quadratic just by seeing them and (-1)^2k was obvious forme

@mertaliozfidan3242

3 жыл бұрын

This comment is one the best ones I have seen I a while

As a former college math professor, I just wanted to weigh in on this channel. You youngsters who are spending time solving these are doing yourself an amazing favor. These are great and fun ways to teach yourself to think of creative and logical solutions that will pay you dividends your entire life. There aren’t too many things you can do now that you will look back on when you are in your 70’s and say “That was a good use of time” but these problems are one of them. The presentation is so good and they have a nice range of problems. I wish I had something this nice when I was a teenager. What a great gift MindYourDecisions has given you!

@thenixaless7493

5 жыл бұрын

I gor 0.11 ^42

@beentherelovedit9150

5 жыл бұрын

You are right, these vids are like intellectual breaks from regular life.

@agassidr

5 жыл бұрын

Thanks for weighing in Sir...

@VikeingBlade

4 жыл бұрын

...When I'm 70 I think I'll be more glad for the moments I lived in than the math problems I solved. (I do love math though.)

@alexj7731

4 жыл бұрын

Ok boomer

Of course I am one year late, but oh man, this is one example why math is gorgeous :D

I am 65 years old. I watch the videos on this channel at night when I can't sleep. I usually only get a partial solution but I got this one totally. I appreciate this channel. I feel it keeps me sharp for my accounting work and my woodworking side business

“How can we solve this problem? Well imagine that you are stuck on this problem” No need to imagine then

Another way to solve: (x^2-7x+11)^(x^2-13x+42)=1 log|x^2-7x+11|^(x^2-13x+42)=0 (x^2-13x+42)log|x^2-7x+11|=0 then,x^2-13x+42=0 or log|x^2-7x+11|=0 first equation: x^2-13x+42=0 (x-6)(x-7)=0 x=6, 7 second equation: log|x^2-7x+11|=0 x^2-7x+11=1 or -1 x=2, 5, 3, 4 So x=2, 3, 4, 5, 6, 7

@vishaljhaveri6176

5 жыл бұрын

That's a very simple method Kyanos. Even I love to do that.

@MrWarlls

5 жыл бұрын

That's a very elegant way to solve this problem. Il like it.

@sk_4142

5 жыл бұрын

Although the video’s solution was equally ingenious, your solution was much simpler and more intuitive. Thank you

@aq9407

5 жыл бұрын

Why you used -1 because log(-1) isn't 0

@sk_4142

5 жыл бұрын

A Q He took the exponential of both sides and ended up with the absolute value sign of the function equal to 1. And since the exponential function and natural logarithmic function are inverses of each other, they cancel each other and thus you can get rid of the natural log. Next, after he gets rid of the absolute value sign, he can consider 1 or -1. Also, just in case you’re wondering why the absolute value is there in the first place, it is because the logarithmic function is defined only for x > 0, that is, only positive numbers.

When I checked the graph in Desmos, I found out that it included the solutions 3 and 4. The reason why you missed it was because the solutions were merely points at the axis.

Try using log in it! It will be wayyyyy easier that way! That way, RHS will become 0 has log 1 is zero, then easy-peasy calculation

@nick46285

4 жыл бұрын

no, if you use log you don't get 6 solutions you only get 4 of them

@engjayah

4 жыл бұрын

You will get all 6 roots Cos log value is the modules of any number which always has two + or - value

@kishore2486

3 жыл бұрын

@Gonçalo Fernandes great

@udith

3 жыл бұрын

You can this without using log also

@rapid13

3 жыл бұрын

Log was my first thought too.

I really enjoyed this problem as an algebra teacher. The first solution that came to my mind was when the exponent was 0.

@KARTIKEYA007

5 жыл бұрын

I am a student who last studied maths a few years ago and all I could think of was 1^y = 1 solution

@pimbe2511

4 жыл бұрын

I got x^0 or 1^x but couldnt think of -1

@msksha6379

4 жыл бұрын

@@pimbe2511 I'm a doctor, and I got similar thoughts

@Saoshi_

4 жыл бұрын

Yeah and I was just thinking to solve x² - 13x + 42 = 0 and you got it

@b.b.b.b.b.b.b

4 жыл бұрын

I am a math teacher's daughter and I found all :)))

My goodness I actually knew how to do this, that's a first

@games267

3 жыл бұрын

I also

@udith

3 жыл бұрын

I got only 4

@joda7697

3 жыл бұрын

@@udith exactly, who knows if there are non-trivial solutions Solving the base to equal 1 and the exponent to be 0 might not give the complete solution set!

@cyndb6303

3 жыл бұрын

I sadly only equated the bases. But I tried.

@jessicajohn6004

3 жыл бұрын

I got the solutions correct that is 7 but I got 6 to be correct too

You are one of the best useful channels of KZread in my opinion Thnx for giving these types of questions these questions fecinate me about know more things

I got the 6 or 7 immediately but did not think of the other two possible solutions. Great stuff.

It's amazing how simple and straightforward the solution is when you just take what is known and connect the dots!

I got 4 just from the thumbnail, but as soon as you mentioned there are 2 more solutions I paused to figure them out. Took a bit, I'll admit, but it was satisfying!

My main man, the TI-92, solved for all 6! Pretty good for a 26-year old graphing calculator.

Case in point wolfram alpha is the best.

@Abhisruta

5 жыл бұрын

No, the human is the grand solver...

@sairamniranjan3908

5 жыл бұрын

You bet it is!

@Abc-np5eo

5 жыл бұрын

There is one equation that wolfram alpha can't solve

@DarthStrider

5 жыл бұрын

There are many things wolfram alpha cant solve but it is the best solver available. Unless of course you kidnap a maths professor.

@sonalidasgupta3562

5 жыл бұрын

@@DarthStrider ....on second thoughts that kidnapped prof + Alpha would be best 😆

That was one of the coolest sulotions I've ever seen for an algebra question....thanks mindyourdesicion

I feel like I had a problem similar to this on a competition where we had an average of 4 minutes to solve each problem (but the questions were overall easier). I forgot about the -1 case, but I remembered it when I was doing the problem for this video!

Thank you Presh! I have a math exam today and I was able to solve this! I have a major confidence boost now😁

@HackedPC

2 жыл бұрын

Ju

This was amazing! I’ve learned the 3rd case (Yes!). The factorial joke at the end is top notch 👏

@xchomphk.9788

Жыл бұрын

i agree this is indeed amazing factorial

Nice trick We know that 2*3*4*5*6*7 = 7! What a nice way

@user-db7rd8fo1w

5 жыл бұрын

oh thanks

@soelinhtet6064

5 жыл бұрын

wowowwwooo…………………… thanks

@AhmadRizviYT

5 жыл бұрын

there is variable too bruh

@Nylspider

4 жыл бұрын

@@AhmadRizviYT lol no U misunderstood the joke

@KrishnaYadav-zt7jp

4 жыл бұрын

Ohhhhh! 7 factorial!! 😅😁

THIS WAS GOOD, THANKS FOR MAKING THIS CHANNEL

I just came across this and thought it was very interesting. I graphed the function and perhaps when you do this kind of problem you could also show the graph. I enjoy seeing it visually.

2,5,6,7 and spent 3 min. And forgotten case3

@emresucu7587

5 жыл бұрын

You forgot the chance that bottom can be (-1) and top can be even number

@j2011j2015j

5 жыл бұрын

Same here then when he said there is 6 solutions I paused it and took me anither 5 minutes to try the negative

@sam3oq980

5 жыл бұрын

3 and 4 work as well because the bottom is -1 and the top is even.

@MichaelRothwell1

4 жыл бұрын

Me too. Only got the last 2 solutions after he insisted there were 6. A great problem, and good video.

@brookehappyeveryday

4 жыл бұрын

me too

It took me 3 minutes to get 2, 5, 6, and 7 without pen and paper. I thought I nailed this one, but incomplete. 🙁

@harrysvensson2610

5 жыл бұрын

mission failed we'll get em next time

@billy.7113

5 жыл бұрын

@@harrysvensson2610 Incomplete mission *≠* Failed mission

@JohnLeePettimoreIII

5 жыл бұрын

Sounds like you did what I did: y^0 1^y And forgot about the third case. That's ok, old dogs can remember old tricks just as we can learn the new ones. 😃

@sonpham3438

5 жыл бұрын

WiseOldDude, same. I was like: set the exponent to 0 and base to 1. Then factor them and bam I‘m done. Haven‘t thought of (-1)^2k thing though

@michak8029

5 жыл бұрын

Did exactly the same..

I was so proud of getting the 4 solutions by myself but i didnt think about the case 3. Nice one man..

I'm in year 10 and I was so close!. I got 7 and 6 by knowing that any number to the power of zero is equal to one, and using simplifying as a way to solve for x, but I never knew the two other methods. thanks for teaching me!

Very interesting! My first attempt was to express the equation E as e^ln(E) to get rid of the exponent and solve E as a zero product, but using ln reduced the interval from ]-∞; +∞[ to [0; +∞[ This is a great reminder that whenever you're getting some function involved into your calculations, you shoud make sure that the domain of the function you're using entirely covers the domain of the base equation

Yeah I missed case 3. Really appreciate this sort of content!

Extremely eye opening thank you i got the four first answers but didn’t think of the two latter solutions.

This one was actually pretty easy. Edit: Damn it, I missed the third case!

@suddeneevee9441

2 жыл бұрын

Same, (-1)^2k is a weird way to say 1. But still valid nevertheless.

I missed the (-1)^y case but got it as soon as you said there were 6 solutions. Wolfram aplha now misses this one too!

I took the natural log of the equation and that allowed me to find 2,5,6,7 but it missed 3 and 4. I think it is because my method assumes that the argument of the natural log is positive.

@jayvyas1827

5 жыл бұрын

yeah i did that too..

@wanyinleung912

5 жыл бұрын

This is probably the method the program used, that's why they missed 2 solutions.

@e11eohe11e

5 жыл бұрын

I did log too. Surprised it wasn't in the video.

@skj84

5 жыл бұрын

Same here, but in my defense I did that mentally sitting on a bus to work.. Guess I could have figured it with a pen and paper

@abhijitsingh3744

5 жыл бұрын

x=3,4 are not missed if you take the natural log and don't restrict yourself to real numbers. (x^2 - 13x + 42) ln(x^2 - 7x + 11) = ln(1) = 0 + i2πk Let x^2 - 13x + 42 = 2k and ln(x^2 - 7x + 11) = iπ. This occurs when x^2 - 7x + 11 = -1 --> x=3,4.

This problem was challenging and interesting. Thanks!

Neat. I did think to consider the cases. Thank you.

I found 6,7 and 2,5 from cases 1,2 in 30 seconds and i thought i was right.... but i forgot case 3... I figured it out the moment you said that many computers only find 4 solutions.....

@himanshujorwal9715

4 жыл бұрын

Same

Checked it with my TI Nspire CX CAS, it found all the solutions too.

Amazing solution.Thank you very much!

Wonderful. I am a math lover and really loved the solution!

I almost got tricked by your sound saying 7 is the answer of multiplying all of the solutions. When you look carefully, the real answer should be 7! instead of 7 and the sentence missed a full stop

@Rekko82

5 жыл бұрын

He forgot the full stop. So it means that he is yelling that the answer is 7!!!! It is important for him us to remember that the answer is SEVEN!!!! Not 7! - it is SEVEN!

@skylardeslypere9909

5 жыл бұрын

That's literally his intention. He wanted to trick everyone by making people like you ( =dumb) think the ! is the full stop of the sentence

@RExN6900

5 жыл бұрын

@@Rekko82 this '!' at the end of a number means the factorial of the number

@braisfernandez5687

5 жыл бұрын

@@RExN6900 He knows

@Rekko82

5 жыл бұрын

@@RExN6900 No, it isn't! If you use ! as a factorial, you still need a full stop to finish the sentence. Well, you don't need - you even start your sentences without any capitals.

When seen the question I did know imediately that there are 2 cases. But I missed the 3rd one. Thank's for this

I only had the two cases in the beginning and then you mentioned that there were six solutions and I immediately was like oh yeah there’s -1 as well🤦🏼‍♀️

Unfortunately you are wrong about 3 and 4 x^2-13x+42 equals 12 for x = 3, but (-1)^(9-39+42) is not the same as (-1)^12, cause to do this step you need this a^(b+c) = a^b*a^c, which for real numbers is correct only when a>0 Also for one problem you can't use two different definitions of same a^b and problem explicetly stated real number so there can not be any ambiguity.

@user-hz8ul4xc1z

3 жыл бұрын

He didnt use that formula a^(b+c), because these are parameters, but u can use numbers so (-1)^(9-39+42) equals to (-1)^12 The formula could be used like if x^(x^2-5x+6) = x^(x^2) * x^(6-5x)

@AndreyGuild

3 жыл бұрын

@@user-hz8ul4xc1z no you can't, given the definition of functions and operators same as sqrt(a*b) != sqrt(a)*sqrt(b), you can't do that willy-nilly even if those are specific numbers

@user-hz8ul4xc1z

3 жыл бұрын

@@AndreyGuild well, not exaclty. The number you are doing sqrt at has to be >= 0, but when it comes to ^ function, it is not necessary, you can do like 5^(-9)

@VillagerJeff

3 жыл бұрын

Even if you had to separate them it would work. (-1)^(9-39+42)=[(-1)^9][(-1)^-39][(-1)^42]=(-1)(1/-1)(1)=(-1)(-1)(1)=1

One such question also came in JEE Main 2016, there they asked sum of all possible values of x and gladly only one sum was matching in the options. It was put under the category of complex numbers. I got it right in the 2nd attempt.😉 👍

yes... and no. If you define x as a real number, it implies that the power notation is just a shortcut for exp((x²-13x+42)ln(x²-7x+11)). Therefore x²-7x+11 can't be a negative number by definition.

@martinmach553

2 жыл бұрын

Who says that ln(x^2 - 7x + 11) needs to be a real number? Take the complex logarithm

@camembertdalembert6323

2 жыл бұрын

​@@martinmach553 and so what ?

@martinmach553

2 жыл бұрын

Huh?

Solving by my own, I only stated the case of exponent equals to zero! Thank you master! It is interesting to learn math the way you do!!!!

Great video! I took all the three cases and answer matched.

Go to Desmos, graph the equation, replace x with another variable like n, create a slider for the variable, set slider step to 1, and slide it over all six solutions shown in the video.

before watching: x^2-13x+42=0 or x^2-7x+11=1 resolts: 6 or 7 or 5 or 2

@aizadzulqarnaen6655

5 жыл бұрын

Here u go 🏆

INDIA CAT 2020, SLOT 1 | Had to solve under 2 minutes, thank god I am your regular viewer! Thanks always:)

Very nice video!! I got it part of this problem because I stopped in case 2. Case 3 I didn't see coming so, I got 4 solutions.

Well I got confused when I ignored "!" and found the product to be 5040😂😂

@yousuf_w1

3 жыл бұрын

this symbol ! means to multiply all numbers under 7 which is 2,3,4,5,6,7

@hellsowner8513

3 жыл бұрын

Me too!

@DatBoi_TheGudBIAS

3 жыл бұрын

Well, now UK how much is 7!

I got a feeling that the only reason why this video was made was because of the joke in the end lol. Cracked me so well

I got the first two cases. Once you mention 6 solutions I had the "aha!" moment and got the 2 solutions from the 3rd case.

The task is to solve for all REAL numbers x, so exponentiation operation b^n is possible only if b > 0. That is why the correct answers are 2, 5, 6, 7 despite the fact that 3, 4 satisfies the equation, however, this operation is not defined. Correct me if I'm wrong.

@heliogen5959

2 жыл бұрын

Are you disputing that 3 and 4 are real numbers? You’re reading too far into it.

Thanks for the puzzle, Presh! The formula will be 1 only when the base has absolute value 1 or the exponent is 0. Since the exponent factors into (x-6)(x-7), we have {6,7} as two solutions. For the base, we have to find roots of (x^2-7*x+11)=1 instead, which is (x^2-7*x+10), which is (x-5)(x-2), giving us two more solutions, {2,5}. Our remaining solutions must have abs(base) = 1 and be raised to an exponent that winds them around to 1. Since we only seek real values for x, our remaining solutions must have base=-1 and an even exponent. So we find roots of (x^2-7*x+11)=-1 which is (x^2-7*x+12) which is (x-3)(x-4), and both substitutions give us even exponents, which gives our remaining solutions {3,4} So in summary we have x in {2,3,4,5,6,7}. Cute.

I accidentally figured out case 3 when I messed up my algebra figuring out case 1.

@tsabotium2211

3 жыл бұрын

"there are no accidents" -master oogway

@pianobear7491

3 жыл бұрын

Task failed successfully!

Very beautiful,very simple.thank u

This was a crazy one that i recognized , and for a while i feel like being on the top of the world.

A similar question has been asked in JEE MAINS 2016.

@jaydani1996

5 жыл бұрын

Yea i just solved it

@aryanlohani622

5 жыл бұрын

It was easy

@tushargola4938

5 жыл бұрын

Yas

@Arpit.singh.

5 жыл бұрын

Yes.... but that question has answer -3 and 10

@rashipplaha1203

5 жыл бұрын

Thats what

For the joke, I started typing 7! into my calculator. Then I noticed that was the joke

Happy to see that the pure Python CAS, SymPy, reports all solutions with a straightforward call to `solve`.

Use log e(ln) on both sides to get two cases, each having 2 real answers, first condition gives 6,7 and the second gives 2,5 and 3,4.

Watching these mathemathical problems and their solutions are like watching Val Valentino/masked magician revealing many magic's secrets. We look for overly complicated solutions only to find out that the solutions are quite simple. Not to mention the misdirections on the problems.

I got 2,3,4,5,6,7 before watching. EDIT: For once I did it right.

@artursanti3276

5 жыл бұрын

X2

@DrakoonLP

5 жыл бұрын

David Nettey same

@Abhisruta

5 жыл бұрын

Good

@davidissel7980

5 жыл бұрын

@David Nettey same for me. I thought I did pretty well until I watched the rest of the video.

@spicydan1582

5 жыл бұрын

[(x-3)*(x-4)-1]^(x-6)(x-7)=1, so either make (x-3)*(x-4)-1 =1, or (x-6)(x-7)=0. therefore I got 2,5,6,7, but 3 and 4 are right too

Today I had a math contest with a similar question. And I could solve it because this video! Thanks a lot

Thank you so much for your effort

On taking logarithm both side ( x^2 - 13x + 42 )*log( x^2 - 7x + 11 ) = 0 On simplifying either x^2 - 13x + 42 = 0 Or log( x^2 - 7x + 11 ) = 0 In other words Either x^2 - 13x + 42 = 0 Or x^2 - 7x + 11 = 1 On solving both quadratic equation, we get solution as x = 2,5,6,7 So four possible answers

@jocabulous

5 жыл бұрын

No

This type of question was asked in IIT JEE entrace exam for engineering colleges in India. The time alloted for each question is 2 minutes. A total of 90 question equally distributed among Maths, Physics and Chemistry for 3 hours.

@ferhatakbulut6572

4 жыл бұрын

ok?

Beautiful solution!

Amazing work

This is the first time I answered a question on this channel without getting anything wrong or missing a possibility :D I'm improving

I missed Case 3 and found 4 solutions...

@sirius9436

4 жыл бұрын

Same

@xxnotmuchxx

4 жыл бұрын

I found 10 solutions

@sunitasharma2536

4 жыл бұрын

Same I did:)

@FoxSlyme

4 жыл бұрын

@@xxnotmuchxx how?

@xxnotmuchxx

4 жыл бұрын

@@FoxSlyme oh nvm, i got 69 solutions

I am not specialised in maths, but I love it. And these videos make me love maths even more

*Amazing trick and explanation*

WolframAlpha to other Math tools: amateurs bprp: believe in math, not in wolframalpha Wolframalpha has left the chat

This is actually easy question for those who are preparing for JEE exams.

@akshatnagar7074

5 жыл бұрын

Yeah, most of these are. It took my class group around 15 min to solve that Gaokao functions question.

@akshatnagar7074

5 жыл бұрын

Did I spell that correctly?

@pranjalpravesh9857

5 жыл бұрын

@@ashutoshmishra7429 jee mains is even very easy as compared to jee advanced.. 😂

@RameshKnowledgeIndex

4 жыл бұрын

They gave 30min per question. We solve such questions in 1min/question rate.

@vijayawasthi70

4 жыл бұрын

yup

Alternatively apply natural log on both sides and ln |x^2 - 7x + 11| has 2 + or - values From which you get the additional 2 roots setting x^2 - 7x + 11 = -1

The joke at the end instantly made me laugh pretty hard

"So, how can we solve this problem? Well, imagine that you're stuck on this problem..." Check!

I got the solutions 2,5,6,7, didn't thought about solutions 3,4

@Stereo4102

4 жыл бұрын

The third case ((-1) ^ 2k = 1) First, convert the result of the base equation in negative: x² - 7x + 11= -1 and turn it into x² - 7x + 11 + 1 = 0 = x² - 7x + 12 = (x - 3)(x - 4) so the solutions are 3 and 4 and substitute the x of the exponent with the solutions to find out that its result is an even number, since no matter whether the number is positive or negative, if it's elevated to an even exponent, the result is always positive.

You are doing a great job.

This is the One of the first type of questions I solved for JEE prep in 11th class

At first this was kinda scary. But after like 2 secs I realized that the exponent should be equal to 0. Ridiculous.

@XGD5layer

4 жыл бұрын

The base being 1 leads to answers too

That's 7-factorial......... My brain went ouch,why didn't I thought bout that after the solutions came out. 😚😚😆

Mygad! I've just known the last case right now😅 thankkk youuuu!!!

Thanks a bunch for such wonderful videos. They really worth watching and sharing. You've changed my attitude towards Math 🙂😃

That last question was really a good trick... I thought it's 7 but actually it's 7! Lol

@prakharrai8819

4 жыл бұрын

It is not 7! . That was a joke🤣 actually 7! =5040.is it the answer?🤨

Technically, it should uave been "You might not believe it, but the answer is 7!."

@artursanti3276

5 жыл бұрын

It's the same... like what? The dot? Xd

@user-vn7ce5ig1z

5 жыл бұрын

Yes, the period. All of his other sentences had proper punctuation (and "own!" at 1:44), so if that was supposed to be a factorial symbol (math needs unique symbols instead of recycling punctuation ¬_¬), then it would have been a hanging sentence without terminating punctuation.

@harshranjan8526

5 жыл бұрын

You are right, but as you know, he wanted to trick us, it was ok.🙂🙂🙂

@Rekko82

5 жыл бұрын

THE ANSWER IS SEVEN!!!!!

@AlexTheUruguayan

5 жыл бұрын

Exactly, lots of riddles rely on improper punctuation as the key to the mystery.

I got 7 & 6 and 5 & 2 very quickly by solving for x^2 -13x + 42 = 0 and x^2 - 7x +11 = 1. However, I initially forgot about solving for x^2 - 7x +11 = -1 (and checking the exponent is an even number) which yields 3 & 4. In all, and incredible set of binomials to produce such a neat result with all the integers from 2 to 7 inclusive giving the right answer. Oh yes, and that little numeric pun at the end (if I can call it that), is almost too clever.

Excellent solution 👍

Ohh, I have found only 4 solutions, the final case is very interesting 7!

@prakharrai8819

4 жыл бұрын

It's not 7!🤨.........if it was 7! then answer would be 5040.don't you think😏

@thebrownintheblack8835

3 жыл бұрын

@PrakharRai I don't know if you're joking or not

@user-lu9qk5ki1q

3 жыл бұрын

Ммм, вершки и корешки

@HerrHoldem

3 жыл бұрын

@@user-lu9qk5ki1q "наши" повсюду

There was a similar question in JEE MAINS or BITSAT exam.

Wow! This was mind blowing!!..

I genuinely expected this to be some insane calculation, but the moment I saw 1^y=1 it all clicked and I was able of doing it