Let y=sqrt(13-x^2), and we change these two equations into: x^2+y^2=13 x^3+y^3=19 Let s=x+y and p=xy, and we have: s^2-2p=13 s^3-3ps=19 Plug p=(s^2-13)/2 into the second equation, and we have: s^3-39s+38=0 This can be solved easily. s_1=1, s_2=(-1+sqrt(152))/2, s_3=(-1-sqrt(152))/2. Note that s=x+y and its max is sqrt(2)*sqrt(13) and thus s_2 and s_3 are rejected. Therefore s=1 and p=-6. That is: x+y=1 xy=-6. And y>=0. This gives us x=-2.
Let [13-x^2]!/2= [19-x^3]^1/3 = y. Then, x^2+y^2 = 13 and x^3+y^3=19. Let x=y=a and xy=b. Then, a^2-2b=13 > b = 1/2(a^2-13). Again, a[a^2-3b]=19.> a[a^2-3/2(a^2-13)] =19 > a^3-39a+38=0 > a=1 > b = -6. Thus, x-6/x=1 > x^2-x-6 =0 > x=-2,3. But x=3 is not a valid solution. So, x=-2.
Θετω 19-χ^3=y^3 και εχω (13-χ^2)^(1/2)=y 13-χ^2=y^2 και εχω το συστημα χ^3+ y^3=19 και χ^2+y^2=13 απο τη λυση του συστηματος και θετοντας χ+ y=α και χy=β στο συνολο Ζ εχω λυση χ=-2. (Η λυση χ=3 απορριπτεται).
The left member is a semicircle, and the right member is decreasing. Looking at the x and y intercepts, you will see only one solution exists. That solution must be negative, between -sqrt(13) and zero. This is where I'd look at integers: -3, -2, -1. Yep, x=-2 is the only real solution.
Horrible method that puts you in a bad situation to solve Hexic polynomial. You have not shown that the remaining quartic has 4 imaginary solutions so that you can disregard it. Also only x=-2 works. You have obtained an extraneous solution with the other one.
Let y=sqrt(13-x^2), and we change these two equations into:
x^2+y^2=13
x^3+y^3=19
Let s=x+y and p=xy, and we have:
s^2-2p=13
s^3-3ps=19
Plug p=(s^2-13)/2 into the second equation, and we have:
s^3-39s+38=0
This can be solved easily.
s_1=1, s_2=(-1+sqrt(152))/2, s_3=(-1-sqrt(152))/2.
Note that s=x+y and its max is sqrt(2)*sqrt(13) and thus s_2 and s_3 are rejected.
Therefore s=1 and p=-6. That is:
x+y=1
xy=-6.
And y>=0.
This gives us x=-2.
let a = (13 - x^2)^1/2 , b = (19 - x^3)^1/3 (a,b > 0) then a = b , a^2 = 13 - x^2 , b^3 = 19 - x^3
=> a^2 + x^2 = (a + x)^2 - 2ax = 13 , a^3 + x^3 = (a + x)^3 - 3ax(a + x) = 19
let p = a + x, q = ax
=> p^2 - 2q =13 ---(1) p^3 - 3pq = 19 ---(2)
p*(1) - (2) => pq = 13p - 19, putting this to (2)
=> p^3 - 3(13p - 19) = 19 => p^3 - 39p + 38 = (p - 1)(p^2 + p - 38) = 0
(case p = 1) => q = (p^2 - 13)/2 = -6, a,x are roots of t^2 - t - 6 = (t -3)(t + 2) = 0
since a > 0, (a,x) = (3, -2) => x = -2
(case p^2 + p - 38 = 0) a,x are roots of t^2 -pt + q =0
determinant(D) = p^2 - 4q = p^2 - 4*(p^2 - 13)/2 = -p^2 + 26 p = (-1 ± 3√17)/2, p^2 = 38 - p
D = -p^2 + 26 = (p -38) + 26 = p -12 = (-1 ± 3√17)/2 -12 = (-25 ± 3√17)/2 < 0
no real solution in this case.
that was a tough one. Great problem and solution.
Let [13-x^2]!/2= [19-x^3]^1/3 = y. Then, x^2+y^2 = 13 and x^3+y^3=19. Let x=y=a and xy=b. Then, a^2-2b=13 > b = 1/2(a^2-13). Again, a[a^2-3b]=19.> a[a^2-3/2(a^2-13)] =19 > a^3-39a+38=0 > a=1 > b = -6. Thus, x-6/x=1 > x^2-x-6 =0 > x=-2,3. But x=3 is not a valid solution. So, x=-2.
Θετω 19-χ^3=y^3 και εχω (13-χ^2)^(1/2)=y 13-χ^2=y^2 και εχω το συστημα χ^3+ y^3=19 και χ^2+y^2=13 απο τη λυση του συστηματος και θετοντας χ+ y=α και χy=β στο συνολο Ζ εχω λυση χ=-2. (Η λυση χ=3 απορριπτεται).
The left member is a semicircle, and the right member is decreasing. Looking at the x and y intercepts, you will see only one solution exists. That solution must be negative, between -sqrt(13) and zero. This is where I'd look at integers: -3, -2, -1. Yep, x=-2 is the only real solution.
X=-2 ; ....
Other solution
Only one real solution x=-2
The given equation equivalent with
[(13 - x^2)^1)2]^6 = [(19 -x^3)^1/3]^6
(13 - x^2)^3 = (19 - x^3)^2
2197 - 507 x^2 + 39 x^4 - x^6 = 361 - 38 x^3 + x^6 or
2 x^6 - 39 x^4 - 38 x^3 + 507 x^2 - 1836 = 0 or
( x +2 )( x - 3)(2 x^4 + 2 x^3 - 25 x^2 - 51 x + 306) = 0.
So x = - 2 and x = 3 . .
Horrible method that puts you in a bad situation to solve Hexic polynomial. You have not shown that the remaining quartic has 4 imaginary solutions so that you can disregard it. Also only x=-2 works. You have obtained an extraneous solution with the other one.
(13)^2 ➖, (x^4)^2={169 ➖ x^4}= 165 10^10^65^1 2^52^5^1^1 1^12^1 2^1 (x ➖ 2x+1) (x^3 )^2➖ (19)^2 ➖ (x^3)^2={x^9 ➖ 361} ➖ x^9={352 ➖ x^9}= 343 10^30^43^1 10^30^1^1 10^5^6 2^5^5^3^2 1^1^13^2 (x ➖ 3x+2).