Math Olympiad | Can you solve given System of Equations ? | 95% Failed to Solve!

[@kareolaussen819]

Beautiful solution❤! Just one comment; you don't have to perform the tedious long division to find the quadratic factor: You know that x+y+z=6 and xyz=4, and have found (say) z =1. Hence x+y=5, xy=4, and x, y must be the roots of t^2-5t+4=0.

[@julietavakian5380]

I did not understand: How did you go from equations(1), (4), (6) to the equation t^3 -6t^2 +9t-4=0 ???? Please, if possible, give me the answer to my question. Thank you.

[@khajabakhangal6134]

Me too

[@aalsii]

If p, q and r are roots of the cubic equation ax³+bx²+cx+d=0 then we have the following relationships among the roots of the cubic equations as follows:- 1) p+q+r= -(b/a) 2) pq+qr+rp= c/a 3) pqr= -d/a and thus the above equation can be written as x³-(p+q+r)x²+(pq+qr+rp)x-pqr=0

[@vijaymaths5483]

Suppose x,y &z are roots of cubic equation in ' t ' then we can write ✍️ t^3 - ( sum of roots)t^2 + ( sum of roots two at a time)t - products of roots = 0 t^3 - ( x+y+z)t^2 + ( xy + yz + zx)t - xyz = 0

[@Rocio62154]

SENSATIONAL!

[@ganeshdas3174]

(x, y, z) = (4, 1 , 1), (1 , 4 , 1) (1, 1,4) will be the real solutions

[@raghvendrasingh1289]

Great From first two equations y z + z x+x y = 9 squaring last equation x+y+z +2 { (√x y)+√(y z)+√(z x) } = 16 √xy+√yz+√zx = 5 squaring again 9 + 2√xyz × 4 = 25 xyz = 4 x,y,z are roots of equation t^3-6t^2+9t-4 = 0 , t= 1,1,4 (x,y,z) = (1,1,4) or (1,4,1) or (4,1,1)

[@akhmadfausi1067]

Mantab . . .

[@ANUDHAWAL]

What happens if the "hit and trial" method does not work? Please explain. Will this method work, if x+y+z=7?

[@kareolaussen819]

In principle yes. But one may easily end up with a 3rd degree equation which is a mess to solve (no integer roots). And worse, with roots x,y,z which cannot be square roots of (even complex) numbers.

[@Quest3669]

X;y; z >0 gir real solns Hence any combo of (4: 1; 1) is o.k

[@Moodeng-r2e]

x=1 y=1 z=4

[@sandeeptanpure8486]

How do u obtain equation t^3-6t^2+9t-4 . Pl explain

[@SGuerra]

Utilize as relações de Girard para polinômios.

[@vijaymaths5483]

If you want cubic equation in ' t ' and x, y & z are be the roots of that cubic equation then we can write ✍️ t^3 - ( x+y+z)t^2+( xy+yz +zx)t - xyz = 0 where x+y+z are sum of roots xy + yz + zx are the sum of roots two at a time and ' xyz ' product of roots

[@ANUDHAWAL]

@@vijaymaths5483 Can this method be used to solve the equations; x+y+z=29, x^2+y^2+z^2=353, x^1/2 +y^1/2 +z^1/2=9?

[@kareolaussen819]

@@ANUDHAWAL Let a=√x, b=√y, c=√z. We want to construct a third degree equation with roots a,b,c. Such an equation is obviously (t-a)(t-b)(t-c) = t^3 - (a+b+c)*t^2 + (ab+bc+ca)*t - abc = 0. We thus need the coefficients A = (a+b+c), B = (ab+bc+ca) and C = abc to construct the equation. A is already given. By squaring A we find A^2 = A_2 + 2B, where A_2 = (a^2+b^2+c^2) is given. Hence B = (A^2-A_2)/2. By squaring A_2 we find A_2^2 = A_4 + 2 B_2, where A_4 = (a^4+b^4+c^4) is given, and B_2 = (ab)^2 + (bc)^2 + (ca)^2 can be computed: B_2 = (A_2^2 - A_4)/2. Finally, by squaring the expression for B we find B^2 = B_2 +2 A*C. I.e, C = (B^2-B_2)/(2A). For your example we have A=9, A_2 = 29, A_4 = 353. Hence B=(81-29)/2=26, B_2 = (29^2-353)/2 = 244, C = (26^2-244)/18 =24. Hence a, b, c must be the roots of t^3 - 9 t^2 + 26 t - 24 = (t-2)(t-3)(t-4). Hence, {a, b, c} = {2, 3, 4} and {x, y, z} = {4, 9, 16}.

[@ANUDHAWAL]

@@kareolaussen819 The solution is (x=4; y=9; z=16). How do you make it so complicated, yet don't get the correct solution?

[@ilafya]

Why all this ,, we have 18= 1+1+16=1^2+1^2+4^2 and the solution are {1,1 4}

[@alexchan4226]

no real solution.