High School Entrance Exam | Only 9% of students got this math question correct

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[@Sqrt.Infinity]

As a person with eyes I solved it and got x = 2 within 10 seconds

[@himanshutilwani10]

Same 😂

[@madelineveggie3931]

x=2 by inspection is one solution. period. This is ridiculous

[@jeanlemire2681]

The obvious answer is x = 2. As can be seen, 2^2 = 4 and 4 + 2 = 6. If I rewrite the equation as 2^x + x - 6 = 0, I can plot the equation y = 2^x + x -6 and I quickly see that there is one real root at x = 2 and there is no complex root.

[@htmlhttpswwwgoo]

More systemically, we can simply specify the range of x into [1, 3) under assumption that x is real. Then quickly find out surveying the natural numbers in [1, 3). Sure, most of people can find out x=2 without systemic approach.

[@BartBuzz]

I find it hard to believe that high school students are introduced to the Lambert W function.

[@TweetyBlu]

I highly doubt about this. This is advanced algebra that is studied at the uni.

[@kosterix123]

Not to my knowledge.

[@2012tulio]

In Egypt we studied it at high school as a way how to solve something like x^x = 5 that can't be solved by factoring .

[@kosterix123]

@@2012tulio interesting - how did that go?

[@BartBuzz]

@@2012tulio High school math has progressed since back in my day. Good to see!

[@X00000370]

Very clever and I enjoyed the practice (even though by inspection x=2 would work nicely on a timed test). That said the Lambert W-Function is very handy and should be in every math major's "tool box".

[@fubaralakbar6800]

You don't need W for this--however it is good that you explain it, because if you had a problem like 2^x+x=5, now you need W. Fortunately, the problem here can be factored. Factor out x for x(2^(x-1)+1=6 Now factor 6 into 2 and 3, guess and check. You find that for 2^(x-1)+1, a value of 2 yields 3--the other factor of 6. So x=2.

[@dieterkohler2206]

The problem is that EVERY eample he picks end up with the W function....

[@Segalmed]

OK one trivial solution is x=2. I'd have to check, whether there are further complex solutions.

[@안진섭-b7w]

램버트함수 안쓰는문제 제시해줌안되나 며칠에 한번 머리 쓴다생각하고 구독했는데 램버트함수문제 나오면 진이 쏙빠짐 이사람 인도사람임? 그렇지않고서는 자꾸 지수+미지수 문제 내는게 말이안됨 저번에도 램버트문제내서 댓글창 씹창났잖아

[@arthurmatthewf5406]

High schools don't have entrance exams. However, 8th graders do learn exponents. So, if a typical 8th grader can't immediately recognize the answer is '2', perhaps the issue is a fundamental flaw in education. And, where does the 9% come from??? Teaching this insanely complex solution to an 8th grader rather than just teaching to employ simple inductive/deductive reasoning will put the child 'off' of math forever. I taught University math and would never abuse college students this way.

[@MattyS4351]

Not all of them, but mine did It's a technical and secondary school, the exam covered 60% of my total score, and the other 40% was covered by my high school grades (the latter was low xd)

[@GeorgeAlexanderOz]

Yeah, great, let's multiply by 2 to the power of 6, but don't ask why.

[@zavianwilson6070]

I like following along... But the teaching philosophy applied in some of his videos are questionable... Learning shouldn't be a quick tricks... Pulling rabbits out of hats! Manipulation is a critical skill needed to master Algeria... not unexplained tricks

[@ruschein]

@@zavianwilson6070 Algeria is a country in North Africa.

[@montyhall2805]

Actually, what he did is fairly common. That is, look at the equation and see what would be applicable (you can toss out that it's linear, etc) and it looks like lambert W is applicable. He's showing how to manipulate the equation to get it into W form. That is when presented with a problem you bring to bear what you know of the "appearance" of what's given to you and then you pull out your bag of tricks to see if you can solve the problem. Being able to recognize what's applicable takes experience and watching how people solve problem just adds to your arsenal of available techniques.

[@elloco1996]

This is idiotic and certainly not material for a high school entrance exam.

[@vicebubble1088]

The function is increasing therefore it has almost once solution. If we replace X with 2 and it verifies the equation, we can stop.

[@Roarshark12]

I liked the second W() trick on the right side of the equation

[@NothingMaster]

Half way through the exercise, your English accent took a beeline for your native language and became almost incomprehensible.

[@irenehartlmayr8369]

This fellows pronunciation is irritating for any musical ear.I cannot understand why this chain cannot find people who speak proper English,to explain Mathematics. Apart from the irritating lilt,I am sick of hearing " naytural " instead of " natural "......

[@davidcarter8932]

Absurd solution. The obvious answer is 2, easily found in certainly not more than the 10 seconds it took sqrt. You have got to be public sector. No one in the private sector would waste 13 minutes doing something that should take 10 seconds.

[@MattyS4351]

Debes buscar soluciones complejas en ejercicios que te exigan encontrar todos los valores posibles de "x"

[@black_eagle]

This might be the silliest math channel on youtube. The solution is 2 by inspection, no need for the lambert W function, which pre-high school students don't know anyway. Ridiculous.

[@dieterkohler2206]

Will you ever stop presenting problems that end in the W function? Do you think that future high school students EVEN KNOW the W function? (Hint: the answer is NO)

[@kosterix123]

Isn’t it obvious that 2 fits the bill?

[@RealQinnMalloryu4]

{2x+2x ➖ }+{x+x➖ }={4x^2+x^2 }=4x^4 (x ➖ 4x+4)

[@AaditiWakude]

X=2 2^2+2=6

[@yuvalmagen100]

beautiful

[@EdwinaTS]

What is lambda function?

[@ChrisHinton1967]

Lambert W function.

[@abdellahabdellah-e1g]

شرح رائع و مبسط

[@GAMERISEDPERSON]

Wow!

[@Misha-g3b]

2.

[@gameknowledgeandit8934]

why not use quadratic formula

[@vijaylaxmigupta7224]

I was wondering the same thing

[@anjajohansson258]

It's not quadratic function, so you can't use the quadratic formula. It reads 2^x, not x^2.

[@anjajohansson258]

I appreciate that one has to show the proof. However, this is very much a special case, as the 6 happens to fit perfectly. A similar solution where 6 is replaced with 7 doesn't give a proper answer. Or would you count a W-function as a final response in those cases?

[@bobbyheffley4955]

​@anjajohansson258 if you replace 6 with 11, x=3

[@anjajohansson258]

@@bobbyheffley4955 yes, I know, because it's also a "nice" number, since 2^3+3 = 11. So what about 7? I'm sure there's a non-integer number that fits the criteria, I just don't know how to solve for it. When trying to put together a formula for solving that distribution of 11 over the two positions, you end up with the original problem.

[@prollysine]

we get , 2^6*ln2=(6-x)ln2*e^((x-6)*ln2) , 2^6=2^2+2^4 , 2^2*ln2*e^(ln2)^4 , 4*ln2*e^(4*ln2) , 4*ln2*e^(4*ln2)=(6-x)ln2*e^((x-6)*ln2) , --> , 4*ln2=(6-x)ln2 , 4=6-x , x=6-4 , x=2 , with W( 2^6*ln2) = 2.772588722 , (6-x)*ln2=2.772588722 , --> , x=6-2.772588722/ln2 , --> , x=2 , OK ,