Solve this to get into Cambridge

[@pradeepsekar]

I like the fact that you shared the longer general method to solve a problem such as this one instead of just the shorter more elegant solution that would have looked more appealing but would not have taught anything of value to the math student leaving them wondering how to pull solutions from a hat and if they need to be able to do so to succeed at math!!

[@Ninja20704]

The shortcut method actually works for a more genral type of nested sqrt problem. If you have sqrt[x +/- 2sqrt(y)], it simplifies to sqrt(a) +/- sqrt(b) if a+b=x and a*b=y. This trick even works if complex numbers get involved. For example to find the principle sqrt of i, we can do sqrt(i)=sqrt(0+i)=sqrt[0+sqrt(-1)] =sqrt[0+2sqrt(-1/4)]. Now think what 2 numbers add to 0 and multiply to -1/4. We can guess and check that 1/2 and -1/2 would be the answer. So, sqrt(i)=sqrt(1/2)+sqrt(-1/2) =1/sqrt(2) + i/sqrt(2)

[@hansduran9462]

Thank u so much! I've been looking for this general way to deal with these.

[@MathFromAlphaToOmega]

Interestingly, it turns out that sqrt(a+bsqrt(c)) can be simplified like this precisely when a²-b²c is a square. There are some criteria for simplifying expressions involving cube roots, but they get much more complicated. Ramanujan found a lot of interesting identities like that - it's called "denesting radicals".

[@kapilchhabria1727]

this is very rudimentary question when studying simplification of radicals, and it warms my heart that none of the pretend AI platforms were able to answer it.

[@JoshuaStorm-zi1wy]

Pretend ?

[@kapilchhabria1727]

@@JoshuaStorm-zi1wy not to get bogged into the semantics of things, but for most parts, these language models are also curating results from prior web searches that are catalogued. it is effectively a very well indexed database if one were to be crudely reductive. essential the concern i am attempting to echo is of the bobbleheads on news networks who complain about jobs being lost to AI. to that, i say, not yet, not for a while.

[@matejdragicevic2067]

I agree, if society forgets how to use their actually brain, it might not end well.

[@JoshuaStorm-zi1wy]

If you do not leverage it to your advantage you will be replaced by someone who does. They will be quicker, more accurate, and more efficient. "AI will not replace you, a person using AI will." @@kapilchhabria1727

[@jongyon7192p]

​@@JoshuaStorm-zi1wy*Real* AI platforms (jk)

[@jasimmathsandphysics]

-1+sqrt(2) surprisingly an easy version for this type of question

[@prnk1729]

3 can be written as (√2)^2+1^2 and then we can apply a^2-2ab+b^2 to get the given surd to be equal to √((√2-1)^2)

[@Misteribel]

For sqrt(a - sqrt(b)), you can use the equivalence equation of sqrt((a + c)/2) - sqrt((a - c)/2), with c = sqrt(a² - b). Set 2sqrt(2) to sqrt(8), and it becomes a trivial exercise.

[@Tiqerboy]

I didn't see the trick, and I solved it EXACTLY as Preshed solved it. It's not often my solution lines up 100% with Presh's but it did in this case!

[@CurlJohnson]

This specific question is taught to students of 8th standard in my country.

[@User-q5m2x]

What country bro

[@tuguldurtumee5059]

I was taught this in 8th grade too. And I’m from Mongolia

[@whyamiwastingmytimeonthis]

Pff, we learn this in kindergarten, your country is so behind

[@aididdat1749]

@@whyamiwastingmytimeonthis I was born with this answer.

[@kelumo7981]

That's not how mathematics is taught buddy😂If schools were teaching kids problems,there's an inexhaustible bunch of problems and noone can ever go through all of them....in mathematics you are taught principles that assist you to tackle a particular set of problems...those principles cab be general enough to cover a wide array of topics

[@craftsmanwoodturner]

I really hope they only asked this of applicants to study maths! Imagine a History applicant being faced with this...

[@leif1075]

You don't think they could solve this..even with having done similar practice problems?

[@Misteribel]

This is a simple high-school or secondary school problem. In most countries that I know of, you learn the basics of square roots in your first or second year. Most 14 year olds will readily solve this. I think it's harder for (most) adults simply because they haven't had the need to use math for such a long time, they just forget. So, any history student should totally be able to answer this.

[@pierreabbat6157]

The problem said that a and b are real numbers. It should have said that they are rational numbers; there's an infinite number of ways of representing √2-1 as a+b√2 where a and b are real numbers.

[@NoahWillCrow]

...How?

[@person1082]

b=0 a=sqrt(3-2sqrt(2))

[@rajkumarroul7306]

Exactly even i noticed it and was confused ;the and would be (a,b)=((1-k ) √2 - 1 , k) where k is any real number

[@dubya5626]

...Because real numbers include irrational numbers. You can wiggle 'a' and 'b' an infinite number of ways to make the equation true. For one example, imagine if we were to equate 3 with a^2 (at 1:25 in the video). In other words, a=✓3. (that is, a=sqrt(3)) Then solve for 'b', which will also be irrational. Real numbers also can be expressed using algebraic numbers, i.e. a certain set of polynomials with integer coefficients. In this manner, they may themselves be denoted as expressions. So let's make our solution as simple as possible and save time: a=✓(3-2✓2) b=0 Indeed, this is how I would have answered the question, though with reservation. Regarding the aspect of there being infinite solutions... Real numbers can also be identified using an infinite number of digits after a decimal point. In this way you can see that there are infinite solutions... So for just ONE example (of an infinite amount of alternative sets available) of infinite series, take the following possible values of 'a' and 'b': a: {-1,(-2.1414...),(-3.2828...),...} (That is {(-1-0✓2),(-1-1✓2),(-1-2✓2),(-1-3✓2),...} And for b: b: {1,2,3,4...} Again, this is but one set of an infinite number of sets available. From the solution suggested in this video, add any number to 'a', and then subtract same from 'b'. The equation will hold true. Therefore, an infinite number of solutions, as original comment stated. Rational numbers on the other hand must be able to be expressed as an integer or as a fraction having an integer numerator and denominator.

[@dubya5626]

Woops. Left and came back amidst that long post. I see others have already answered. Anyway, I hope the long description helps those who may require it.. and kudos to the others who posted! Hope the content creator reads this and corrects the video. Make sure you up vote the original comment so he (and all) can see the original comment!!!

[@mike1024.]

As soon as the full question was said instead of what was in the thumbnail, it seemed like a pretty easy question to me. Certainly not that easy if you haven't done anything involving equating coefficients in the past though!

[@Inquisitive_Nomad]

We in India do it in class 8 or 9, I don't remember but it was one of these two

[@mike1024.]

@@Inquisitive_Nomad Do you use the same way Presh did it?

[@69Gigantosaur]

​@@mike1024.i am also from india and this is the method i know: √3-2√2 = √x - √y Squaring both sides 3 - 2√2 = x + y - 2√xy x+y = 3 -2√xy = -2√2 √xy = √2 xy = 2 From this x=2 and and y = 1 √x = √2 √y = 1 √x -√y = √2 -1

[@RyukenBlack]

​@@Inquisitive_Nomadi dont think so i am from india, i mean at 8-9 we know what is square and what is underroot but dont know how to solve this type of specific questions where the expression become perfect square and etc i learned this technique in 11th

[@RikMaxSpeed]

I really love the short trick solution at the end. Also fun that the current AIs can’t solve it.

[@user-oh2kt8lf6g]

3:54 The way I solved it initially. It is a good practice to see if there is a full square before turning to equations.

[@AntonAdelson]

Smart!

[@Blade.5786]

The -2√2 immediately looked like a -2ab from (a-b)², and I proceeded from there. It was pretty cool how they all cancelled out.

[@AA-100]

Wouldnt it be better to write the final solution as -1 + Sqrt(2) since the question asked it to be in the form a + b*sqrt(2)

[@aroundandround]

0:20 As strictly phrased, a reasonable answer could be a=0, b=sqrt(1.5 - sqrt(2)). Maybe the question intended to specify rational, not real.

[@driksarkar6675]

I agree, the step at 1:24 fails for real numbers but works for rational numbers.

[@angeluomo]

Wolfram Alpha had no problem simplifying this expression.

[@Gabitza379]

We could also use this method: √(a±√b) = √[(a+c)/2] ± √[(a-c)/2] ; c=√(a²-b) So a=3; b =2√2 =√8; c=√(9-8) =1 √(4/2) - √(2/2) = √2 - 1

[@bigmviperplayz253]

But don't we also get √2+1

[@Gabitza379]

@@bigmviperplayz253 Well, in our case, we use the "-" from the "±". If it was √(3 +2√2), then we would use the "+" and get √2+1 It's my fault, I should have written this instead: √(a+√b) = √[(a+c)/2] + √[(a-c)/2] √(a - √b) = √[(a+c)/2] - √[(a-c)/2] ; c=√(a²-b)

[@bigmviperplayz253]

@@Gabitza379 thanks for clearing it up 🙏

[@Gabitza379]

@@bigmviperplayz253 Always happy to :)

[@tambuwalmathsclass]

The shortest way to solve it is to think of two numbers that can be multiplied to get 2 but added to 3. The number are 2 and 1. But we need their square roots as a solution, i.e √2 and √1 then take the difference i.e √2-√1 which is √2 - 1

[@kaninma7237]

I set a + b(root 2) squared equal to the answer and obtained two equations for the two unknowns, determining that a=-1 and b = 1. One equation came from the fact that only one term in the expansion of the square had root 2 in it, showing me that ab = -1.

[@galois6569]

I think the question is actually asking for the solution where a and b and integers (or rationals). as it is written you could just answer b=0 and a=sqrt(3-2sqrt(2))

[@mccannms]

The LLM AIs have had problems doing math. Wolfram Alpha, in addition to just calculating the decimal value of the answer, also gives this solution and the start of the step by step process in the free version.

[@qwang3118]

After getting a^2 + 2b^2 = 3, a*b = -1, it's easy to see that the conjugate sqrt [3 + 2sqrt(2) ] = a - b*sqrt(2) or - [a - b*sqrt(2) ] > 0, so one of them is valid, and the other invalid. Then (a, b) = (1, -1), -(1, 1), (sqrt(2), -1/sqrt(z)), - (sqrt(2), -1/sqrt(z)). 2 of them lead to sqrt(2) - 1 > 0, the other 2 lead to 1 - sqrt(2) < 0, invalid.

[@DasMonitor1]

Uhm, I think the question is actually meant to be "for rational numbers a and b", because with real numbers a = √(3-2√(2)) and b = 0 would be a valid solution in the requested form, as both a and b are real numbers, and the sum √(3-2√(2)) + 0 * √(2) definitely satisfies as a solution to √(3-2√(2))😅.

[@RGP_Maths]

Exactly what I was thinking!

[@rajkumarroul7306]

Exactly even i noticed it and was confused ;the and would be (a,b)=((1-k ) √2 - 1 , k) where k is any real number or rational number also

[@choke7273]

Using (a+b)²=a²+b²+2ab, you could turn it into 1-2root(2)+2, which is (1-root(2))², if yout take the root it becomes ±(1-root2), and it must be positive so root2-1, much simplier

[@rajkumarroul7306]

@MindYourDecisions you have to either restrict a,b to rational(or integers) to obtain the solution you found. The step you did at 1:20 to 1:30 where u equated 3 to a²+2b² and 2ab to -2 is totally wrong for real numbers it is possible only if a,b are rational(or integers)

[@rohitchaoji]

Yeah I was thinking the same thing. They either have to be rational or integers, or it has to be a quantity you can separate out in components like these, such as vectors or complex numbers (but in that case the components are the unit basis vectors in case of vectors, or Re and Im in case of complex numbers).

[@vladimirkaplun5774]

Nested radicals is a well known formula. The canonical method is to consider a=sqrt(3+-2sqrt(2)) and b=sqrt(3+-2sqrt(2)) ab=1,a^2+b^2=6.

[@shiveshpratapsingh3501]

√(3-2√2) the inside expression i.e. 3-2√2 can be written as (√2-1)^2 then the square root and the square will cancel out leaving √2-1. I have a ninth grade side book of mathematics the author is R S Aggrawal and found that problem there . And another mathematics book written by R.D. Sharma these two books contain high kevel questions

[@cronox2585]

This is easy man, I don't know what you are talking about. All you have to do is notice the identity and then it can be solved. It is not arduous at all.

[@pluisjenijn]

Fun! But in the form a + b√2 the answer √2 - 1 is in the wrong form, isn't it? Correct answer is -1 + √2

[@xnick_uy]

Isn't the question stated in a somewhat vague form? Any real number x satisfies x = 0 +[x/√ (2)] * √ (2) so we could have trivially said a = 0, b = x/ √ 2 as a solution... Even more trivial would have been to set a=x and b=0. And asides from that, you can set either a or be to any value you like, and figure the other one to satisfy the relation.

[@patel_pankaj_69]

For any root of surds ; √(a ± 2√b) = √x ± √y. x,y ∈ Q+, Where, √b = Surds & a ∈ Q+ and a² - 4b is the square of rational numbers. √(a ± 2√b) = √x ± √y is the solution of root of surds. Where, a = x + y & b = x * y. Therefore, √(a - 2√b) = √(3 - 2√2) Where a=3 & b=2 So, x and y is the common factor of a & b. a = x + y = 2 + 1. b = x * y = 2 * 1. ∴ x = 2 and y = 1 Therefore solution is √x - √y. √(a - 2√b) = √(3 - 2√2) =√2 - √1 = √2 - 1. ★ That's the Ans. @MindYourDecisions ✌️❤️✨

[@riy0h]

Why compare with unintelligent language models and not mention wolframalpha at the end?

[@devondevon4366]

sqrt 2 - sqrt 1 Using the quadratic formulae x^2 - bx + c=0 x^2 - 3x + 2 2+ 1 =3 2*1 *1=2 sqrt 2 - sqrt 1

[@ARex545]

We can simplify it as: root(root2-root1)² => root2-1 Thus the answer is root2-1

[@ARex545]

Currently studying for CAT, it involves solving many questions on this logic.

[@mohdbilal2]

Exactly, i think this is most easiest way to solve this problem but this is tricky so it may not click to everyone

[@ARex545]

@@mohdbilal2 That is true. I would not have been able to solve this very quickly if I did not have a grasp on this logic beforehand. But, once we understand the logic behind it(completing the square), we can put as many underoots in the question and it will be easily solvable.

[@mohdbilal2]

@@ARex545 💯

[@iceman9678]

Who knew it was so easy to get into Cambridge?!?!

[@denisripley8699]

Always useful to know that root 2 is 1.414

[@Mythraen]

I suspect that's only sometimes useful.

[@denisripley8699]

@@Mythraen I was of the slide rule and Log-Table generation.. I, and most contemporaries) could generally recall all the 4-figure Logs (and anti-Logs) to a 1-2 places. It was nothing exceptional to folks (engineers) back then. Some could get 5 figure logs correct more often than not.

[@Mythraen]

@@denisripley8699 ... I was poking fun because you said "always." A lot of people, a lot of the time, do not need to know the root of 2. Hence: "sometimes."

[@shoaibahmad3259]

Easiest question in my life

[@peterdavis9403]

I mentally remembered sqrt2 ~ 1.414 and that 2sqrt 2 ~2.818 so the sqrt (3-2.818) = sqrt (.1820) then worked it out as .414 and figured it might be (sqrt 2)-1 Tested by "trick method" and was done.

[@pradeepsekar]

I did exactly the same...!! But the more general method that Presh presented was the right one for any math student

[@Mathcentricmind]

Before the 3:54 he is proving the formula, the method finding full square starts after that. 3=sqrt2^2+1^2 and so on.

[@yuresko1349]

I was about to write my own way of solving it, but it turns out it is exactly the quick way of doing that 🤑

[@WarmWeatherGuy]

I put the equation in my calculator and got 0.414213562 and I kick myself for not realizing that is 1-sqrt(2). Instead I worked the problem exactly the way that Presh did. This is very unusual because I rarely get the right answer the same way as done in the video.

[@mike1024.]

I think you meant sqrt(2)-1. ;-)

[@WarmWeatherGuy]

@@mike1024. Yes 🙂

[@1zu87Fuchs]

This one was trivial at first sight and also proofs as a 30 sec. task: it is obvious to view 3 as 2+1, use a binomial formular and instantly wirte down the answer. Challabge us harder. Presh, I love your channel.

[@mhm6421]

As an asian I solved it in 3 seconds (i just did sqrt( (sqrt(2) - 1)^2 ) which just gets out from the sqrt)

[@loez2111]

Türkler asyalı olmuyor ki

[@JaggedFel621]

Eyeballed it in decimal form, got ~.4, surmised it must be -1 + sqrt(2). Fuzzy math wins.

[@Decocoa]

The chad way

[@jirizelinka9056]

Is is possible to solve the problem by an easier way: If x=a+b*sqrt(2) then a^2+2b^2=3, 2ab=-2 (see video solution). Thus (a+b)^2+b^2=a^2+2b^2+2ab=3-2=1. So sum of second powers of two integers is equal to 1, so one of them is equal to 1 and the second is equal to 0. The only possibility for non-zero b and positive x is b=1, a=-1.

[@AKBARCLASSES]

Thank you

[@gtabro1337]

1:29 wait you can do that? I've never seen this trick, how do we know _precisely_ these polynomial members equal the other ones chosen, like is there a factor or we can just pick random ones from the left to equate to others chosen from the right?

[@zlac]

we don't, it's a trick because we know that the solution must be in the form od a+b(root2)

[@mike1024.]

Equating coefficients is a pretty standard procedure for dealing with equations that can be characterized by some sort of extension form, more commonly a variable like x. The lowest level topic I've seen it used is partial fractions (more often when the case involves an irreducible quadratic), which is usually taught in calculus 2 but in theory could be taught in a precalculus class. The reason it works boils down to the definition of a field extension, basically taking some element and its powers, multiplying by coefficients from that field of numbers, and adding the terms together. For example, polynomials are an extension of the real numbers with the variable x. In this case, we are extending the real numbers with the square root of 2. In a field extension, two elements are equal if and only if their corresponding coefficients are equal in the field of numbers where they originated.

[@NovaH00]

Erm this trick is used quite a lot actually, I don't know the English term but my language translation is "matching coefficient".Basically, the sqrt(2) is an irrational number so the only chance that both side are equal is when the coefficient of sqrt(2) are equal (you can't create sqrt(2) by +,-,×,÷ numbers). And then if the coefficient of sqrt(2) are matched then the constant term should be matched too. The "matching coefficient" is used when you know for sure there is none other way to create a coefficient. Like for example: 2x + 4x^2 = ax + bx^2 If a,b are integer you know for sure that the only way that both side are equal is when a = 2 and b = 4 But if a,b are real number, we can still match the coefficient but there are lots of cases and there are infinite sol, which is not practical Case 1: the above case Case 2: Let a = 4x then ax = 4x^2. Matching the "2x" coefficient with bx^2, so b = 2/x (2/x.x^2 = 2x) So a = 4x, b = 2/x (inf sol) Case 3... and so on. Inf cases, some case with inf sol

[@ExplosiveBrohoof]

If a number n lies in the field Q(sqrt(2)), then it must be able to be uniquely written in the form a + b sqrt(2) for rational numbers a and b.

[@SeriyRL]

We had this task when we were at 8 grade.

[@POLARTTYRTM]

Why do I watch these if I don't know or can't comprehend even simple math? This fries my brain.

[@ericfielding668]

I think there is something missing from your problem's constraints. If a and b can be any real numbers, there is a much simpler solution: a = 0 and b = sqrt[ (3/2) - sqrt(2) ]. Each real a will have a corresponding real b forming a solution. Maybe you need to restrict a and b to rational numbers in order to give the problem some level of difficulty.

[@rajkumarroul7306]

Exactly even i noticed it and was confused ;the and would be (a,b)=((1-k ) √2 - 1 , k) where k is any real number or rational number also

[@rohitchaoji]

Is there a proof for why we can separate out the "components" of this equation like that? The solution seems nice and elegant but I'm sure there are some restrictions or some criteria for doing the step of equating the components of the equation like that.

[@MichaelRothwell1]

Yes there is, as long as we start out by looking for solutions of the form a+b√2 where a, b are rational (in fact they will be integers or half integers). We can then equate the rational coefficients of √2 and the rational terms, and later reject the irrational solutions for a and b. Unfortunately this was not clear from the video.

[@Rknife]

I figured it out in my head!

[@MarieAnne.]

I did too, pretty much using the method shown at the end.

[@earthbind83]

1:25 How did you know you could split your equation into two equations like that?

[@nna7yk]

IS IT EASY?? In math, never say that!! Of course we know >>>> | 4cos((75/2)°)cos((195/2)°) | = √(2) - 1

[@GeoffryGifari]

can we generalize the last bit? if a square root is equal to a sum of terms, then whatever's inside that square root must be a perfect square?

[@MarieAnne.]

For the solution shown, instead of solving √(3 − 2√2) = a + b√2 and getting a quartic equation, it's a bit simpler to solve √(3 − 2√2) = √a − √b and getting a quadratic equation: √(3 − 2√2) = √a − √b 3 − 2√2 = a + b − 2√(ab) a + b = 3, ab = 2 → b = 2/a a + 2/a = 3 a² + 2 = 3a a² − 3a + 2 = 0 (a − 1) (a − 2) = 0 a = 1 → b = 2/1 = 2 → √(3 − 2√2) = √1 − √2 = 1 − √2 < 0 → not valid a = 2 → b = 2/2 = 1 → √(3 − 2√2) = √2 − √1 = *√2 − 1* > 0 → OK ChatGPT solution shown in video is triply wrong. First it says that the solution to √(a−2√b) = √p − √q where p and q are roots of x² − 2ax + (a²-b²) = 0 But the correct quadratic equation is x² − ax + b = 0 with roots p and q where p > q. Second, when solving, it claims that a = 2, b = 2. But clearly a = 3. Third, if it plugged in a = 2 and b = 2 into its own quadratic equation, it would get x² − 4x = 0, which has solutions p = 4, q = 0. Or if it had plugged in correct values a = 3 and b = 2, it would get x² − 6x + 5 = 0, which has solutions p = 5, q = 1 I have no idea how it came up with p = 2 + √2, q = 2 − √2 On the other hand, if we plug in correct values a = 3, b = 2 into correct equation x² − ax + b = 0, we get: x² − 3x + 2 = 0 (x − 2) (x − 1) = 0 p = 2, q = 1 √(a − 2√b) = √p − √q → √(3 − 2√2) = √2 − √1 = √2 − 1

[@handanyldzhan9232]

[a + b * sqrt(2)]^2 = a^2 + 2b^2 + 2ab * sqrt(2) = 3 - 2 * sqrt(2) a^2 + 2b^2 = 3 2ab = -2 b = -1/a Trivial answer = sqrt(2)-1

[@coolliger]

Can someone explain 1:24? How did he just set those equal to each other?

[@driksarkar6675]

I think the question’s supposed to say that a and b are both rational numbers or integers; this doesn’t work if they’re allowed to be any real number, and a and b aren’t unique in that case.

[@gavindeane3670]

He set them equal to each other by inspection. If 3 - 2√2 is equal to (some stuff) + (other stuff)√2 then we can immediately see that if (some stuff) is 3 then (other stuff) must be -2.

[@zinc_magnesium]

Believe it or not I had a question using this exact trick on my homework last week.

[@daakudaddy5453]

I can just write a + b root(2) = root( 3 - 2 root(2) ) + 0.root(2) Both a and b are real numbers. Done. a

[@yaduk7710]

I did EXACTLY the same method(1st one) as in the video accidentally. From equating the coefficients to seeing that 3-2rt2 > 0. Freaky

[@jeroenvanrensen9505]

Aren’t there an infinite amount of solutions? If a = sqrt(3 - 2sqrt(2)) - b sqrt(2), then it is a valid solution it seems to me.

[@adipy8912]

BlackPenRedPen is a great channel

[@user-gs4cq3uk2b]

∴How to remove double radicals √(a+b)±√(ab))=√a±√b in this case a=2 b=1

[@PeteZam]

wolfram alpha had no problem finding the form -1+sqrt(2). looks like bard and chapGPT need to step up their game, because that is a trivial problem for wolfram alpha

[@philipkudrna5643]

The same old trick every time. 5 sec: the solution is sqrt(2)-1. The easier (and more intuitive) way to solve this would have been to show that 3-2sqrt(2) can be written as 2-2*sqrt(2)*1+1, which is equal to (sqrt(2)^2-2*sqrt(2)*1+1^2 or simply (sqrt(2)-1)^2. From: (a-b)^2, where a=sqrt(2) and b=1. But I guess the chosen method would work with every number, also ones that do not lead to a „pretty“ solution like in this case.

[@MrMasterGamer0]

I just use the identity: sqrt(a+/-sqrt(b))=sqrt((a+sqrt(a^2-b))/2)+/-sqrt((a-sqrt(a^2-b))

[@pooroldpedro]

gpt4 got this wrong first attempt, so I said "are you sure" and it tried again and solved it (with simultaneous equations)

[@WhiteGandalfs]

Would be a better representation of the video if the WHOLE request would have gone into the thumbnail. I pretty much like to solve your problems on my own BEFORE viewing the video, then first watching the end to compare my results, and only after that eventually (if my results didn't match) watching more details, usually going backwards through the video - until i grasp the clue. I this case, that was impossible due to the fact that a necessary part of he request was kept hidden in the first minute of the video.

[@arnavverma2461]

This is like the easiest question possible

[@GrifGrey]

whats 1 + 1 would be easier (using basic arithmetic, that is)

[@arnavverma2461]

@@GrifGrey no ig 0+0 would be easier

[@GrifGrey]

@@arnavverma2461you right

[@MichaelZankel]

3:54 this is a similar way but it doesn’t work(I’ll be using S for square root to make this shorter): 1. S(3-2s(2))= 2. S(1+((s(2))^2) - 2*s(2)) = 3. S(1-s(2))^2 = 1-s(2) but that’s less than 1 Ohh, at the end it’s + or - that answer in line 3 but you didn’t specify this part of it in the video

[@MathsScienceandHinduism]

nothing tricky about this. the term inside the root is clearly expressible as a whole square

[@Trust_the_brain]

man, just simplify a + b * sqrt(2) to x or something it makes it way easier to understand and the a + b * sqrt(2) part can be done later on.

[@erikaz1590]

"A.I., how do I solve this equation?" ["Beeeeep. I don't know. MindYourDecisions hasn't made a youtube video about it yet. Beep Boop."]

[@khangdao8119]

In Vietnam, we learn to solve problems like this at grade 9

[@anurupsil8216]

I asked both chatgpt and bard and both were able to solve it. For chatgpt I used bing because bing uses "gpt4 testing version". Bing solved it in the first chance but I had to correct the bard in some lines so the bard got solved in the second chance. But they didn't consider the primary roots so they gave me both negative and positive results

[@oguzkaaninanc1658]

We learned this in 9th class

[@ramunasstulga8264]

Everyone in the world have learnt this in 8th grade 💀

[@f.r.y5857]

√(a+b±2√(ab)) = √a ± √b This property helpful.

[@NiharTripathy_Dew_Lord]

We study how to solve such questions at a very early age, in class 8th and 9th in India.

[@bobh6728]

Is √ 2 - 1 really in the form asked in the question which was a + b√ 2? Just being picky.

[@Wmann]

Blackpenredpen!!!

[@ExplosiveBrohoof]

This question confuses me. Doesn't a=sqrt(3 - 2sqrt(2)), b=0 work?

[@rexanguis214]

I’m new to math, why does he set it equal to (a plus b(root2))……why would that work, god bless

[@Dominus_Potatus]

... this is 9th grade? The idea is facrorize the inside root 2 only has 1 pair = 1 × 2 find pair that if summed is equal to 3 So the answer is sqrt(2) - sqrt(1)

[@Dominus_Potatus]

But I understand that it is a memorization and math need to understand how to solve

[@ratandmonkey2982]

I don't think I would have gotten into Cambridge.

[@robertsolovy69]

If you have to know this before you show up to school, what's the point of paying them for an education

[@shabnambano7291]

Being an Indian. I solved it just after seeing the thumbnail.

[@nedmerrill5705]

nice one.

[@bhartiyekunal]

I am from Bharat and this question i read in class 9th.Your method is so complex,I can solve easily.

[@jayeshkumar3861]

I took nearly 0.75 seconds to solve this. Anyways.... U r my role model in math.

[@lpslpslpslpslpslps]

This is a weird cumbersome solution. Why not just: sqrt(3 - 2sqrt(2)) = sqrt (2 - 2*sqrt(2) + 1) = sqrt ((sqrt(2) - 1)^2) = sqrt(2) - 1

[@tjitseberkey9484]

I solved it in my head so much easier, at least I think so

[@nripdave673]

I sent one intresting algebra question suggestion on your email.... Can anyone tell after how much time it will be posted in upcoming video.....

[@ABhaim]

I tried to approximate and I got 0.4 - no calculator or sheet of paper

[@ntlake]

Uhm... you know, you could just say a = sqrt(3 - 2*sqrt(2)) - sqrt(2) and b = 1. They're real numbers, aren't they?

[@kam7056]

💡

[@teambellavsteamalice]

Once you know the trick you can do these kind of problems in a few seconds. 😁

[@ai2657]

√2-1

[@chinnayyakorrai8523]

7th-8th class question in India

[@SASA_maxillo]

Bruh, i wish we took the same 😭😭

[@rise4562]

Nahh 9