Math Olympiad | A Nice Algebra Problem | How to solve for "a" and "b" in this problem ?
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Nice solution, but looks like it is from some ancient world, where no formula for quadratic equation roots is known yet ). It could have been done in a more straightforward way: after we got equation a+b+1=0 we get a= -1-b from here and then substitute a in Eq #2 with this. We get then b^2 +b+1=133. From here we get b^2 + b -132=0. And we just find roots using standard way, via discriminant.
Thanks for hard work
Good, but too much redundant works! Once you've got your (eq.3) a+b=-1, then just substituting it into either of the given eqns (eq.1 or .2) yields the same results simply.
Agreed. The substitution style works well for more complex problems but here it's unnecessary.
The segment of this content is to an elementary student
Not for the experts
Agree with this. Another way to make it more efficient (though less than above, but you dont have to solve the quadratic) is once you have arrived at a+b=-1 and ab=-132, simply consider the possible factors of 132, and deduce the (obvious) solutions.
@@davidbagg9289це можна зробити просто в голові без запису.
Якщо зробити, як Ви пропонуєте.
Я вже рішив таку задачу просто закривши очі.
В России за такое извращение учителя будут ругаться на учеников)) метод подстановки, ну вырази а через b и всё...каждый раз на этом канале удивляюсь. Простейшие задачи называют олимпиадными и решаются максимально криво. С одной стороны, весело за таким наблюдать. С другой стороны, это очень печально..
Даже не решал уравнений.
а*2 = 133 + b
Единственный квадрат, удовлетворяющий этому равенству,
144 = а*2
Какое длинное и муторное решение. Гораздо проще и короче решить через систему уравнений, выразив одну неизвестную через другую
Видно у них такой стереотип - каждый шаг разжёвывать до мелочей.
Aquí tenemos un ejemplo de la diferencia entre eficacia y eficiencia.
La resolución fué eficaz pues se llegó al resultado correcto; pero no fue eficiente pues no utilizó los conocimientos que le permitirían llegar con menos recursos y muchísimo más rápido.
It is do easy equat 1 - equatio2 : ( a- b) * ( a+ b) = -( a-b,) therefore a=b cancell or a+b =,-1. B= - a-1 conclude
a^2 +a +1= 133 conclude a =11 or -12. B= -12 or 11 thats all
In depth explanation for newbs or 'I forgot school math' guys. Very nice. Ty.
For a complete solution of the system you have to notice first the symmetry in a,b which shows that a= b is a possible case.
Hence, there also two complex solutions together with solution pairs (a , b) ,(b,a) .
a^2 - b = 133 = b^2 - a, so a^2 - b^2 = b - a, so (a+b)(a-b) = -1(a-b), so a+b = -1, since a NOT = b.
So, a and b are the 2 roots of the symmetric quadratic (-x-1)^2 - x = 133 or x^2 + 2x + 1 - x = 133 or x^2 + x -132 = 0
So. x = (-1 +/- 23)/2 = (11, -12) = (a,b) or (b,a) by symmetry.
I can't do anything, but I keep an eye on these types of problems and the solution comes quickly, in this case, as the values are different, it gets a little complicated, but the trick is that one value is positive and the other negative.
a = - 12
b = + 11
144 - 11 = 133
121 - (-12) = 133
Bingo from Brazil
Симпатичное решение интересного задания!
The way the professor explained is impressive
Math Olympiad? Perhaps for grade school. 2b or not 2b. That is the question.
a^2 - b = b^2 -a
a^2 + a = b^2 +b
We are looking for two points on the parabola y = x^2 +x with same y-value. They must be symmetric to the lowest point with x coordinate -0.5
So a = -0.5 - c and b = -0.5 + c
(-0-5 - c)^2 - (-0.5 + c) = 133
0.25 + c +c^2 +0.5 - c = 133
c^2 = 132.25
c = root (13225/100) = root (25*529/100) = 5 * 23 / 10 = 11.5 (or -11.5)
So a = -12 and b = 11 (or opposite)
Từ hệ phương trình đã cho, suy ra :
a*2 - b=b*2 - a
tương đương
a*2 - b*2= - (a - b)
(a+b)(a - b)= -(a - b)
Chia hai vế cho (a - b),ta được:
a+b= -1 hay
-b=a+1
Thay vào a*2 - b=133 ta có
a*2 +(a+1)=133
a*2 +a - 132=0
Giải phương trình trên thì đạt được hai cặp nghiệm:
a=11 ;b= -12
a= -12 ;b=11
Good
1+root533over two for all of them a and b
Instead of adding both equation, aimply use the first equation with the 3rd one using elimination and sovle the quadratic to get a=11 or -12.
I am not sure if my solution is acceptable. My initial thoughts were
a^2 + b = 133 -- (1)
b^2 + a = 133 -- (2)
(1) = a^2 + b = 121 + 12
Therefore a = 11 or -11 b = 12
However equation (2) shows that the vice versa. And hence one must be a negative therefore a = -11, b = 12 combinations would work for both equations.
Nice solution
Thank you very much!!
Used usual replacement, some approximation to ease calculation and got -12 and 11
11 and - 12
11 sec and all on my mind
One of the numbers is -12, another one 11. Took 5 secs.
👍👏
Life is already hard, please don't make it harder...
Great solution!!
Это для дебилов? В 1972 году меня моя математичка в школе за такое извращение над уравнением поставила бы угол. Привет от образовения в СССР!!!
Once a+b+1 is determined to be zero, simply substitute b = -a−1 in first equation to get
a²−(-a−1) = 133
a²+a−132 = 0
(a+12)(a−11) = 0
(a, b) ∈ {(-12, 11), (11, -12)}.
Only one check is needed because the system is symmetric.
Correction: first two equations were slightly wrong.
You made the wrong substitution: should be a2+a+1=133; a2+a-132=0 ; but the results are OK.
@@Ginkobil50 Yes, first two equations are incorrect. Updated.
(a²-b) / (b²-a) = 1
Etc ...
My intuition told me -12 and 11 in about 2 seconds. Didn’t have to watch but it’s probably right
Mathematicien invented i squared to be - 1, why not inventing a letter = to x divided by 0 ?
Thanks.
You only checked the solutions in one equation each.
Where are the other real roots?
another solutions:
(a,b) = (12.0434) [approximations for irrational numbers]
(a,b) = (-11.0434) [approximations for irrational numbers]
like a² - a = 133
i love the "hello" at the beginning
Was gonna write the same. Helps me relax
@@alfonata74same
Yes! I got it!
265 = (2^2 + 1)(2^2 + 7^2)
Use two graphs intersection.
😮
It’s not Olympiad task, too simple
和と差の積ですねわかります
① +② get ③ the same
①+③
→ a^2 + a = 132
a = 11 or -12 → ②
if a =11 then
b ^2 - 11 = 133
b = ±12 → ①
b = 12 (X)
b = -12 (O)
if a = -12 then
b^2 + 12 =133
b = ±11 → ①
b = 11 (O)
b = -11 (X)
Answer：
a =11 and b = -12
or
a = -12 and b = 11
-12 & 11
Attendance link... forgot to post it...pl. send it
a² - b² + a - b = 0
(a - b)(a + b + 1) = 0
a = b => rejected
b = - (a + 1)
a² + a + 1 = 133
a² + a - 132 = 0
a = (-1 ± 23)/2
a = 11 => b = -12
a = -12 => b = 11
11 и -12 в уме подобрал за пару минут. 😅
b = - a - 1 = - ( a + 1 )
a^2 + a + 1 = 133
a^2 + a - 132 = 0
a = { 11 , - 12 }
( a , b ) = { ( 11 , - 12 ) , ( - 12 , 11 ) }
😊🤪👍👋
11,-12
a = +- 12 b = +-11
красиво
Ais 12 B is 11
Difficile
Hi dear teacher where are you from
a=11 b=-12
(-12;11)(11;-12)
At the 154 seconds, you have violated PEMDAS, whereby, you added ((a-b) + (a-b) before multiplying (a+b) (a-b). This changes the answer. Please explain.
He was not using PEMDAS. He was simply factoring out (a-b) from each term.
From the given equation we can infer that a² and b² are close to 133 and that perfect squares near this number are 121 and 144. Hence the value of a and b is either + or - 12 and + or - 11 or vice versa. By suitably substituting these nos. in the given equation we may get the answer in simple way.
Was it necessary to impose the condition a≠b in the question? It is clear that even if such a condition is not there a= b cannot be a solution since the LHS of each equation then becomes zero which is not equal to the RHS, 133.
(-12;11), (11; -12)
Я устно решил, за 15 секунд, методом подбора. А = 11
В = (-12)
なぜ、英語圏の学生は数学が苦手なのか？が、何となく分かったよ(°ω°)
IT'S SORRY YOU'LL WRITE SO MANY LETTERS IF YOU TYPE 12 x 12 = 144 -133 = 11.help me, you are very complicated.
Как долго...
Thanks for wasting 14 minutes to solve a 30-second equation.
I can't imagine following this path on a GRE or an SAT 😂
Пипец как долго, в уме решается за 2 минуту, и выразить а через б их не учат чтоле?
SOLUTİON -->> czcams.com/video/L9F4MOg54Ps/video.html
Das más vueltas que un carrusel...
É LAMENTAVEL VOCÊ ESCREVER TANTA BESTEIRAS 12 X 12 = 144 -133 = 11
Let a = x )))
a = -12 b=11
тупое, длинное решение. Да еще и жутким акцентом, явно индус.
Нормальное. Всё чётко.
долго, нудно, непрактично.