Math Olympiad | A Very Nice Algebra Problem

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Ай бұрын

Math Olympiad | A Very Nice Algebra Problem
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Пікірлер: 77

@danieldepaula6930 Ай бұрын

I think an easier way would be just defining c=sqrt(a) and d=sqrt(b). After that, we have a simple system of sum (S) and product (P) that can be solved by the equation x²-Sx+P=0, where the roots are c and d. Then, just find a=c² and b=d².

@fredericrl3576 20 күн бұрын

Yes 👍

@vladimirroboz 17 күн бұрын

😊😊

@belowtale3414 17 күн бұрын

Ah yes viète

@xynyde0 9 күн бұрын

thats just the same solution, with extra steps...

@alagaika8515 Ай бұрын

Once you have a+b=80, you can substitute a=40-x, b=40+x and enter that into ab=100. Much easier to do without paper, no need for completing the square.

@RandomGuyEdit 7 күн бұрын

You are wrong there is need to make a quadratic equation ,, because there is not any definitions like a,b ∈ R ... So ,, a and b can be imaginary ...

@imprettynein 2 күн бұрын

@@RandomGuyEdit He is not wrong. When a=40-x and b=40+x, the sum a+b is still 80, no matter if it is real or imaginary.

@tvc3mye Күн бұрын

Brilliant idea! This substitution will skip the entire step of applying the quadratic root formula altogether.

@silver6054 Ай бұрын

1) The equations are obviously symmetrical in a and b, so once you have found the first ordered pair you can write down the second. 2) There was a lot of squaring, so it would be worth checking that the answers found are valid 3) You use sqrt(a)*sqrt(b) = sqrt(a*b) without noting any conditions. OK in this case as both a and b are positive

@Kounomura 25 күн бұрын

The two-variable equation can be immediately reduced to a one-variable one, due to the symmetries. Let sqrt(a)= 5-x, sqrt(b) = 5+x. Thus (5+x)(5-x)=10 --> 25-x^2 =10 --> x = sqrt(15) --> a=(5-sqrt(15))^2, b=(5+sqrt(15))^2 or vice versa.

@higenharinson9207 24 күн бұрын

б забыл в квадрат возвести

@Kounomura 22 күн бұрын

@@higenharinson9207Действительно, спасибо

@economicsonline 18 күн бұрын

I watch these videos on double time. Too many easy steps drawn out separately and needlessly. Don't have to show subtracting in two steps. Just do it.

@blue_birb 15 күн бұрын

I did it orally before opening the video and now I feel better about myself

@arthur_p_dent 24 күн бұрын

6:27 at this point, if the previous caluculationsa are all correct, then due to the symmetry of the problem, if a=40+10 sqrt(15) then b must be 40-10 sqrt(15) and vice versa.

@user-il4hm4yi4p 28 күн бұрын

Элементарно же: из второго уравнения выражаем одну из переменых, подставляем в первое, получаем квадратное уравнение относительно корня, отрицательное значение отбрасываем, положительное возводим в квадрат. Подставляем найденое значение во второе уравнение, находим вторую неизвестную.

@user-nj7od8cp3k 16 күн бұрын

В обоих квадратных уравнениях либо а

@TheXJ12 24 күн бұрын

Since √ab is non-zero, a is non-zero and I can multiply the first equation by √a --> (√a)² + √ab = 10√a --> (√a)² + 10 = 10√a --> A² + 10 = 10A then solve A² -10A + 10 = 0

@subhashdeshpande1772 2 күн бұрын

You are working like a coolie !!

@pgbrandon 28 күн бұрын

Nicely done

@starpawsy Ай бұрын

I would square both equations, then substitute the second into the first. Fairly straightforward from then on.

@user-mm2gg1xh1z 18 күн бұрын

Можно решить в 5 строчек если предствить корни из a и b как корни квадратного уравнения x^2 - 10x +10=0

@lightning77125 20 күн бұрын

a = 40+10(sqrt(15)) b = 40-10(sqrt(15)) Square both sides, solve for (a+b) and (ab), use Quad. formula to get answer.

@KPunktFurry 3 күн бұрын

oh i had a mistake during moveing of the equantion and got a negativ d :D but thanks for the right solution! LG K.Furry

@Chiter_haip 3 күн бұрын

То самое оформление второй части огэ:

@yifengxiao 26 күн бұрын

let √a and √b be two roots of a quadratic equation x² -10x+10=0, and solve it. This idea is applicable to any case with the form of sum and product of two variables. For example, you may solve x³ +y³=5, x³y³=6, or a ∛a + ∛b =6, ∛ab =8 in the same way. just as danieldepaula says.

@zerocat888 4 күн бұрын

I solved the same way!

@Sungkookpark11 12 күн бұрын

Good. Easy ones.

@theeternalsw0rd 24 күн бұрын

If the problem is asking for real solutions only, you need to show that 10*sqrt(15) < 40 to prove that 40 - 10*sqrt(15) is positive. To do this, divide both sides by 10 so that sqrt(15) < 4 then square both sides 15 < 16, which is true.

@youandi7142 Ай бұрын

a + b + 2 akar ab = 100 a + b + 20 = 100 a + b = 80 (akar ab)² = 10² ab = 100 (a + b)² = a² + b² + 2ab 640 = a² + b² + 200 a² + b² = 440

@user-ud3cg2vp5s 29 күн бұрын

Nice❤

@davidbrown8763 15 күн бұрын

Thanks for there challenge. Nailed it in one.

@_AGCraft_ 8 күн бұрын

So nice sir

@gheffz 20 күн бұрын

What's this one? Sqrt(ab) = 10 & Sqrt(a) + Sqrt(b) = 7

@KhedrBadr 8 күн бұрын

What is the name of the program you use ?

@mryip06 19 күн бұрын

Recently invented method by someone I forgot let u > 0, √a = 5+u, √b = 5-u √a√b = 25-u² = 10 u = √15 √a = 5+√15 a = 25+10√15+15 = 40+10√15 Similarly, b = 40-10√15

@Sichy_ 2 күн бұрын

is this olympiad? so easy problem i think

@mdahnafalfee196 18 күн бұрын

Is the answer correct? If yes, then why am I not getting the exact result (10) after placing the values of a & b? I'm not even getting closer values. If I put values in √a+√b, I'm getting this answer : 12.634...........!! And If I put values in √ab, I'm getting this answer : 39.81205..........!!! Somebody help plz!!

@user-ow1fs8et9l 16 күн бұрын

Это не олимпиадное задание, мб обычное из учебника со звёздочкой

@lowwl1fe Ай бұрын

Solving scholar system was like it is IMO

@user-bd2ue7vb5l 28 күн бұрын

У нас на олимпиадах и в учебниках ответы были посимпатичнее, а не корень из 15

@user-gi3zl7nv9t 15 күн бұрын

let sqrt(a) = x, sqrt(b) = y then x + y = 10, xy = 10 -> (x, y) = (5 + sqrt(15), 5 - sqrt(15)) or reversed. therefore, (a, b) = (sqrt(5 + sqrt(15)), sqrt(5 - sqrt(15))) that's the easy way.

@tvc3mye Күн бұрын

This answer doesn't seem right. If you really have to pursue this path to solving the equation, it should have been: x = 10 - y xy = 10 → y(10-y) = 10 y² - 10y + 10 = 0 y =½(10 ± √60) = ½(10 ± 2√15) b = y² = ¼(160 ± 40√15) b = 40 ± 10√15 a = 40 ∓ 10√15

@marklevin3236 22 күн бұрын

40+sqrt1500 and 40-sqrt1500

@nestoreleuteriopaivabendo5415 28 күн бұрын

Sorry about the change of subject, but what music is that?

@mathshunter 28 күн бұрын

You can get the name of the music in description.

@nestoreleuteriopaivabendo5415 28 күн бұрын

@@mathshunter I'm not finding it, sorry. Here it just says "Math Olympiad | A Very Nice Algebra Problem Hello My Dear KZfaq Family 😍😍😍 Hope you all are doing well 🥰🥰🥰 If you are enjoying my video about how to solve this math olympiad algebra problem then please Like and Subscribe my channel as it helps me a lot 🙂🙂🙂" But it looks a lot like Yawning Man, I like it!

@mathshunter 27 күн бұрын

The music is parasail, icing and nine lives.

@user-el3yn5gh4p 26 күн бұрын

😂вы гениальный сумасшедший ❕

@shazhu2455 19 күн бұрын

Why don’t you just solve for sqrt(a) and sqrt(b) instead?

@magneticpropulsion2507 Ай бұрын

Plug these answers back into the original equations and doesn't equal 10. ???? Spotted one error in calculations.

@user-fm3no5gm9t 24 күн бұрын

I checked - Everything is right!

@FuturePast2019 22 күн бұрын

Almost works for A=81 B=1

@peternewseterforever 12 күн бұрын

If we work in such values, sqrt(ab) = 10 would be false.

@E.h.a.b 24 күн бұрын

(√a + √b)^2 = (10)^2 a + b + 2 √ab = 100 a + b = 100 - 2 * 10 a + b = 80 ----------> [1] (a - b)^2 = (a + b)^2 - 4 ab (a - b)^2 = (80)^2 - 4 (10)^2 =6400-400 (a - b)^2 = 6000 = 4 * 15 * 100 a - b = +/- 20 √15 ----------> [2] 2 a = 80 +/- 20 √15 Add [1]+[2] a = 40 +/- 10 √15 a = 40 + 10 √15 AND a = 40 - 10 √15 b = 80 - a b = 40 - 10 √15 AND b = 40 + 10 √15

@SergiuCosminViorel 23 күн бұрын

too easy! olympian?! what level?

@mathshunter 23 күн бұрын

Junior level (for grade 8 students)

@malto_only Ай бұрын

I don't like using • as multiplication symbol. Why do start using this after highschool lebelt? I can't think of any logical origin of this dot. All other symbol have a logical origin +-×÷

@GreyJaguar725 Ай бұрын

when x is a variable, otherwise you would get something like x×2x×x² vs. x•2x•x²

@malto_only Ай бұрын

@@GreyJaguar725 you see, x and × are really not that similer, it's possible to diffrenciate. X has higher angles while × has lower angles

@malto_only Ай бұрын

@@GreyJaguar725 you see, x and × are really not that similer, it's possible to diffrenciate. X has higher angles while × has lower angles

@starydinozaurofficial250 Ай бұрын

@@malto_onlyif thats not enough, there is something called the cross product which is also an x, so it would just be too confusing

@peteevans7776 Ай бұрын

You used an and b as variables and then coeficients in the quad equation, bad practice, can confuse novices. You should have used different characters.

@silver6054 Ай бұрын

Yes, my thought. And even earlier, showing the expansion of (a+b)^2 and sqrt(a)*sqrt(b) = sqrt(a*b) although these are a little less confusing

@nasrullahhusnan2289 Ай бұрын

sqrt(a)+sqrt(b)=10 sqrt(ab)=10 --> [sqrt(a)][sqrt(b)]=10 sqrt(a)+sqrt(b)=[sqrt(a)][sqrt(b)] By inspection: --> sqrt(a)=2 and sqrt(b)=2 a=b=4

@tvc3mye Күн бұрын

with your answer, √a+√b will not be 10, which fails to satisfy the 1st equation

@nasrullahhusnan2289 Күн бұрын

@@tvc3mye: Thanks for the corrections. I was misled by the exactly the same value 10 of RHS of the two equations. The correct solution can be found by subsitution. Substitute sqrt(b)=10/sqrt(a) from the 2nd equation to the 1st one, then simplify to get a quadratic equation in x. Then use to find y by using 1st equation to be simpler. Then square x and y. As the two equations are cyclical, interchanging x and y does not change the equation, we can just use one value of x to get y and then interchange what we get to find the other one.