Upon seeing the thumbnail, I did it this way: a² - b = b² - a, a(a+1) = b(b+1). One solution, which we don't want, is a = b. For the other solution, we can write a(a+1) = (-b)(-b-1). This gives a = -b-1. So b² + b + 1 = 43 or b(b+1) = 42. So b = 6 and a = -7 is a solution. By symmetry, a = 6 and b = -7 works. (Again as before, one could say (-b)(-b-1) = 42.) Since the equations are quadratic, there are no more solutions.
Before doing any work, see if it can be quickly solved by assuming the answers are integers and some trial and error. Logical first guess is a=7, which give us b=6. That doesn’t quite work in the second equation, so realize a has to be -7.
If both a,b>0，or a,b0，b0 subtract it from the other equation we get (a+b)(a-b+1)=0 So a=-b or a=1-b The first is ruled out. Substituting the second we will get the answers. Remember to add back the negative sign for b.
... It's also instructive to mention the geometrical side of the system .... namely, 2 parabola's ( 1 horizontal par. opening to the right & 1 vertical par. opening upwards ) intersecting with each other in 4 distinct points, but because due to the restriction A /= B, we only mention 2 intersections, and consider the other 2 as non valid! ... Jan-W
Wow, you took the long route to reach your destination. Transitive property, grouping, followed by a quick factor of the trinomial and you’re done in under 45-seconds.
That's on the RUclips end, not necessarily under the creator's control. The only easy way around it is to tack on a 20 second outro after the problem is finished.
It is very similar to Ramanujan's pair of equations which is y + √x = 7 x + √y = 11 Answers are very simple ( x = 9 , y = 4) But solution is............... 🙄
... Good day, Given: Eq.(1) A^2 - B = 43 & Eq.(2) B^2 - A = 43 & A /= B ... A^2 - B = B^2 - A .... A^2 - B^2 = - A + B ... (A - B)(A + B) = (A - B)(- 1) ... applying (X) * (Y) = (X) * (Z) X = 0 v Y = Z ... A - B = 0 ... A = B (REJECTED) v A + B = - 1 ... then A = - 1 - B & recalling Eq.(1) .... (- B - 1)^2 - B - 43 = 0 ... B^2 + B - 42 = 0 ... (B - 6)(B + 7) = 0 ... B = 6 v B = - 7 .... recalling A = - 1 - B .... B = 6 , A = - 7 v B = - 7 , A = 6 ... finally (A , B) : S = { (- 7 , 6) , (6 , - 7) } ... the same outcomes as yours ... thank you for your presentation ... Jan-W
Upon seeing the thumbnail, I did it this way: a² - b = b² - a, a(a+1) = b(b+1). One solution, which we don't want, is a = b. For the other solution, we can write a(a+1) = (-b)(-b-1). This gives a = -b-1. So b² + b + 1 = 43 or b(b+1) = 42. So b = 6 and a = -7 is a solution. By symmetry, a = 6 and b = -7 works. (Again as before, one could say (-b)(-b-1) = 42.) Since the equations are quadratic, there are no more solutions.
Before doing any work, see if it can be quickly solved by assuming the answers are integers and some trial and error. Logical first guess is a=7, which give us b=6. That doesn’t quite work in the second equation, so realize a has to be -7.
If both a,b>0，or a,b0，b0
subtract it from the other equation we get
(a+b)(a-b+1)=0
So a=-b or a=1-b
The first is ruled out.
Substituting the second we will get the answers. Remember to add back the negative sign for b.
... It's also instructive to mention the geometrical side of the system .... namely, 2 parabola's ( 1 horizontal par. opening to the right & 1 vertical par. opening upwards ) intersecting with each other in 4 distinct points, but because due to the restriction A /= B, we only mention 2 intersections, and consider the other 2 as non valid! ... Jan-W
a^2-b=b^2-a, (a^2-b^2)=b-a, (a-b)(a+b+1)=0, so a+b+1=0, b=-(a+1), a^2+2a+1-a=43, a^2+a-42=0, (a+7)(a-6)=0, a=-7, b=6, or a=6, b=-7.😊
I solve very fast by only mental exercice;
A = + 6
B = - 7
Bingo
From Brazil!
a²-b=43
b²-a=43
a=√(43+b)
b²-√(43+b)=43
b²-43=√(43+b)
(b²-43)²=43+b
b⁴-86b²-b+1806=0
(b²+b-42) (b²-b-43)=0
(b-6) (b+7) ...
[b-((1+√173)/2)] [b-((1-√173)/2)]=0
a≠b → a,b=(6,-7)
Wow, you took the long route to reach your destination. Transitive property, grouping, followed by a quick factor of the trinomial and you’re done in under 45-seconds.
A^2-A-43=0 means A and B being 1/2±sqrt(173/2).
How about showing the problem to its conclusion before overlaying the video with other problems.
That's on the RUclips end, not necessarily under the creator's control. The only easy way around it is to tack on a 20 second outro after the problem is finished.
a≠bなら、どっかでa−bかb−aで割れば解けるようになっているのかな？😊
Yes. Possible.
(6; -7) et (-7; 6)
It is very similar to Ramanujan's pair of equations which is
y + √x = 7
x + √y = 11
Answers are very simple ( x = 9 , y = 4)
But solution is............... 🙄
Exactly...
I solved it by brute force all in my head. I knew using my head for brute force would pay off someday.
a=-7 and b=6
b:6; a:(-)7
-7 , and 6
but how to calculate if 3. condition (a is not b9 is not available?
(a is not b)
a=6, b=-7
A=6 b=-7
I just left this problem unsolved today. And have seen it after coming home.😢😢😢
There's no way this is an olympiad problem!
This is a question for the math olympiad for dummies.
ты можешь блять условия писать в превью
... Good day, Given: Eq.(1) A^2 - B = 43 & Eq.(2) B^2 - A = 43 & A /= B ... A^2 - B = B^2 - A .... A^2 - B^2 = - A + B ... (A - B)(A + B) = (A - B)(- 1) ... applying (X) * (Y) = (X) * (Z) X = 0 v Y = Z ... A - B = 0 ... A = B (REJECTED) v A + B = - 1 ... then A = - 1 - B & recalling Eq.(1) .... (- B - 1)^2 - B - 43 = 0 ... B^2 + B - 42 = 0 ... (B - 6)(B + 7) = 0 ... B = 6 v B = - 7 .... recalling A = - 1 - B .... B = 6 , A = - 7 v B = - 7 , A = 6 ... finally (A , B) : S = { (- 7 , 6) , (6 , - 7) } ... the same outcomes as yours ... thank you for your presentation ... Jan-W