Solve the simultaneous equations x^2+y^2=10 , x^4+y^4=82
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We have to solve the simultaneous equations
x^2 + y^2=10 ...(1)
x^4 + y^4=82 ...(2)
let a = x^2 and b = y^2
This gives
a + b = 10 and a^2 + b^2 = 82
a + b = 10
=> a^2 + b^2 + 2ab = 100
substitute a^2 + b^2 = 82
=>...
(The entire section contains 113 words.)
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x^2+y^2=10 (1)
x^4+y^4=82 (2)
(x^2+y^2)^2= x^4+2x^2y^2+y^4
Since x^2+y^2=10 and x^4+y^4=82 we have:
100=82+2xy^2
then:
2x^2y^2=18
(xy)^2=9
xy=3 xy=-3
from (1)
x^2+y^2+2xy= 10+6=16
(x+y)^2=16 x+y=4 xy=3
being symmetric: x=1 y=3 x=3 y=1
x+y=-4 xy=3 x=-1 y=-3 x=-3 y=-1
or
x^2+2xy+y^2= 10-6=4
(x+y)^2=4 x+y=2 xy=-3 x=3 y=-1 x=-1 y=3
or
(x+y)^2=-4 x+y=-2 xy=-3 x=-3 y=1 x=1 y=-3
So the couples we are searching for are:
(1,3), (3,1),(-3,-1),(-1,-3),(3,-1),(-1,3),(-3,1),(1,-3)
Since the 2nd equation could be written as:
x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2
But x^2 + y^2 = 10 (from the 1st equation),
x^4 + y^4 = 10^2 - 2(xy)^2
82 = 100 - 2(xy)^2
2(xy)^2 = 100 - 82
2(xy)^2 = 18
(xy)^2 = 9
xy = +3 or xy = -3
We'll re-write the 1st equation:
x^2 + y^2 = (x+y)^2 - 2xy
10 = (x+y)^2 - 2xy
For xy = 3, we'll get:
10 = (x+y)^2 - 6
(x+y)^2 = 16
(x+y) = 4 or (x+y) = -4
We'll create the quadratic when we know the sum and the product of the roots:
x^2 - Sx + P = 0
For s = x+y = 4 and P = 3
x^2 - 4x + 3 = 0
x1 = [4 + sqrt(16 - 12)]/2
x1 = (4+2)/2
x = 3 and y = 1
x = 1 and y = 3
For s = x+y = -4 and P = 3
x^2 + 4x + 3 = 0
x1 = [-4 + sqrt(16 - 12)]/2
x = (-4 + 2)/2
x = -1 and y = -3
x = -3 and y = -1
For xy = -3, we'll get:
10 = (x+y)^2 + 6
(x+y)^2 = 4
x + y = 2 or x + y = -2
For s = x+y = 2 and P = -3
x^2 - 2x - 3 = 0
x = [2 + sqrt(4 + 12)]/2
x = (2+4)/2
x = 3 and y = -1
x = -1 and y = 3
For s = x+y = -2 and P = -3
x^2 + 2x - 3 = 0
x = [-2 + sqrt(4 + 12)]/2
x = (-2 + 4)/2
x = 1 and y = -3
x = -3 and y = 1
The solutions of the system of equations are: (3;1);(1;3);(-3;-1);(-1;-3);(3;-1);(-1;3);(1;-3);(-3;1).
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