Math Olympiad | A Nice Algebra Problem | How to solve for 'a' and 'b' in this problem?
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- čas přidán 27. 08. 2024
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Thank You. But after 3:08 it’s possible a little differently. (1) a^2-b=111 ; (2) b^2-a=111 ; (3) a=-b-1 . We substitute (3) in (2) . We get : (4) b^2+b-10*11=0 . Therefore : (5) b1=10 , b2=-11 . ( thanks to Viett ) . We substitute (5) in (3) - we get Your answer.
With respect, Lidiy
I think that there is no need to add eqn1 to eqn 2 . Just use your a+b+1=0 which means that a = -b -1 then substitute a = -b -1 into eqn1 you will get
a = (-b-1) substitute this into eqn 1 and you get
(-b-1)^2 -b = 111 expanding you will get
(b^2 + 2b +1) -b = 111 simplifying gives
b^2+b+1 = 111 => b^2+b+1 -111 =0 which then gives
b^2+b -110 =0 => b^2+11b-10b-110 =0 => b(b+11)-10(b+11)
(b-10)(b+11) =0. so b has two values (10 and - 11)
substitute these values in eqn 2 to get a with two values of -11 and +10
so (a,b) =(-11,10) and (a,b) = (10, -11). Hope this helps
I stopped at equation 3. Very brilliant!
Good job
Thank you very much!!!
Beautiful!🎉
Thank you very much!!!
Or just make a simple substitution to get a quartic equation, use the possible rational zeros theorem to identify possible roots, and do a little synthetic division...
how about bbc basic?
10 sw=1:a=sw:goto 40
20 b=a*a-111:dgu1=b^2/a^2:dgu2=1/a:dgu3=111/a^2:dg=dgu1-dgu2-dgu3:return
40 gosub 20
50 dg1=dg:a1=a:a=a+sw:a2=a:gosub 20:if dg1*dg>0 then 50
print a,b
This is the fourth time you have solved the same problem, just with different values for the right hand side: 111, 133, 41, 33.
You are taking longer each time.
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(-11;10)(10;-11)
A = 24642.
B = 12321.