Math Olympiad | How to solve for "a" and "b" in this problem?

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@gaiatetuya92 2 ай бұрын

a=7のとき左辺は44を越えるので a=1,2,3,4,5,6のどれかしかない。順に与式に入れていけば a=2のときb=8, a=3のときb=5 が出る。この方が大分早い。

@alexanderserov3952 2 ай бұрын

(a+b) ^2 = 44+(b-1)*b Therefore 44+(b-1)*b is a full square. (b-1)*b is an even number and is a product of two consecutive numbers. Therefore b=5 and a=3 or b=8 and a=2. Sorry for my English.

@mustafasibic2954 2 ай бұрын

I have solution 1) Factor: a^2+2ab+b^2=(b+a/2)(1+2a)=44 2) Multiply by 2 (2b+a)(1+2a)=88+a 3) write b in term of a(idk this how to say it) 2b+a= (88+a)/(1+2a) 2b= (88+a)/(1+2a) - a 4b= (176+2a)/(1+2a) - 2a 4b= 1+ 175/(2a+1) - 2a 4) find b values 175=1×175=5×35=25×7 after trying we will find the solutions which is (a,b) = {(2,8),(3,5)}

@SALogics 2 ай бұрын

Very nice trick to solve this type of problems! ❤❤

@билал-ж2к 2 ай бұрын

Спасибо!!! Чудесное решение!!! ❤

@ericjb1030 2 ай бұрын

Case 1 could have been rejected when you found a=0 without computing b.

@pas6295 2 ай бұрын

a^2=9 is s a=3. Then 6b+b=44-9=35.So 7b=35. b=5. Answer ais 3 and b is 5.

@فیروزاهنگری Ай бұрын

If b=1 , a^2+2a×1+1=44 , (a+1)^2=44 , a+1=✓44 , a=5•6 result 0

@juvenalkanani2458 Ай бұрын

Also, the couple (0; 44) verifies the original equation. Therefore it might be acceptable!

@giuseppemalaguti435 2 ай бұрын

a=-b+√(b^2-b+44)..b=-4(a=12,-4)....b=-7(a=17,-3)

@chao541 Ай бұрын

44 is very small. Just try 0 to 6 for a.

@davidshen5916 Ай бұрын

44=A^2+2AB+B=B(2A+1)+((2A)^2-1+1)/4=(2A+1)(B+(2A-1)/4)+1/4, 44*4=(2A+1)(4B+2A-1)+1, (2A+1)(2A+4B-1)=175

@madanmohan1221 2 ай бұрын

please give answer me how to prove it = +-√-1 please🙏

@danikochman1351 2 ай бұрын

(a,b)=(2,8);(3,5);(12,-4);(-13,5);(-4,-4)

@pas6295 2 ай бұрын

a =3 b=5.

@pas6295 2 ай бұрын

Why do much of complications. Induction method is the easiest. Between 1 and 9 on seeing the equation by giving one value to one unknown and get the value of the other in order to satisfy the equation.

@JunedHussain 2 ай бұрын

Positive integer is more than 1-9

@pas6295 2 ай бұрын

@@JunedHussain The very problem is such two unknown but one equation. Since it happened to lenear one you can have multiple answers. Had it been Quadratic equation you get only their roots. But in Lenear one for two unknown to have only one answer you need one more equation.

@JunedHussain 2 ай бұрын

@@pas6295 good

@letsfindthejams3525 2 ай бұрын

From one equation you can not calculate the value of two ends a and b therefore all is wrong

@raj-nq8ke 2 ай бұрын

Comparison part is wrong .

@Chapulin28 2 ай бұрын

Why?

@yanssala2214 2 ай бұрын

Haz supuesto 3 casos pero en realidad son infinitos casos. De otro modo hablando su unica ecuacion no tiene solucion unica al tener dos incognitas. Borre esto de internet y no tupa a los estudiantes. Queridos estudiantes si a=0 entonces b = 44 si a =1 b= 4,333; si a =2 , b =8; si a=3, b =5; si a =4, b =3,111...y asi hasta infinito. El valor de a puede ser cualquier numero real y estos no son contables. No crean en este farsante y estudien matemáticas en serio.