The given system of equations is x² − y² = 40 xy = 99 and we are required to find the value(s) of x + y. First note that we have the _identity_ (x + y)⁴ = (x² − y²)² + 4xy(x + y)² which can be easily derived by rewriting (x + y)⁴ as (x + y)²(x + y)² = ((x − y)² + 4xy))(x + y)². Now let x + y = s and substitute this as well as x² − y² = 40 and xy = 99 in the identity above, then we have s⁴ = 1600 + 396s² s⁴ − 396s² − 1600 = 0 s⁴ − 400s² + 4s² − 1600 = 0 s²(s² − 400) + 4(s² − 400) = 0 (s² − 400)(s² + 4) = 0 s² − 400 = 0 ⋁ s² + 4 = 0 s² = 400 ⋁ s² = −4 s = 20 ⋁ s= −20 ⋁ s = 2i ⋁ s = −2i
Indeed a clever way to get the sum s w/o calculating x and y first if you know this trick. However, in a test situation with time restraints I would prefer a bit clumsy but easier to spot substitution y = 99 / x or y² = 9981 / x² which gives us (x² - 81) (x² +121) = 0.
@@MrGeorge1896 Both are nice solutions! I was even lazier, and calculated d = x - y: (y+d)^2 - y^2 = 40 (y+d) * y = 99 Using the first equation: y = (40-d^2)/2d = 20/d - d/2 x = y + d Putting that into the second eqation: (20/d-d/2 + d)(20/d-d/2) = (20/d+d/2)(20/d-d/2) = 10^2/(d/2)^2 - (d/2)^2 = 99 Let m = (d/2)^2: 100/m - m = 99 or m^2 + 99 m - 100 = 0 m_1/2 = -99/2 +/- sqrt(9801/4+100) = -99/2 +/- sqrt(10201/4) = -99/2 +/- 101/2 = { -100, 1 } m = (d/2)^2 > 0, thus m = 1, d = 2. We already suggested, that 99 = 11 * 9 was one of the sensible integer solution, x = 11, y = 9, but we have to acknowledge, that x = -11, y = -9 is also a solution (from the structure of the problem). The funny thing is, that x = 9i, y = -11i, and their negative counterpart also work! (9i)^2 - (-11i)^2 = -81 - (-121) = 40, (9i)(-11i) = - (-99) = 99, and (-9i)^2 - (11i)^2 = -81 - (-121) = 40, (-9i)(11i) = - (-99) = 99.
@@rainerzufall42 If you are going to work with the difference of x and y as a variable, then it makes sense to work with the sum of x and y as a variable as well. The given system of equations is x² − y² = 40 xy = 99 Using the identities a² − b² = (a + b)(a − b) and (a + b)² − (a − b)² = 4ab this system can be written as (x + y)(x − y) = 40 (x + y)² − (x − y)² = 396 Now, if we let x + y = s x − y = d we get sd = 40 s² − d² = 396 Observe that the structure of this system is similar to that of the original system, except that the variables are now s and d. You can solve this system for s in any way you like. A conventional approach would be to substitute d = 40/s from the first equation into the second equation, then we have s² − (40/s)² = 396 which gives s⁴ − 396s² − 1600 = 0 Note that this is the exact same biquadratic equation in s which I obtained directly by using the identity (x + y)⁴ = (x² − y²)² + 4xy(x + y)² with the original system in x and y.
The given system of equations is
x² − y² = 40
xy = 99
and we are required to find the value(s) of x + y.
First note that we have the _identity_
(x + y)⁴ = (x² − y²)² + 4xy(x + y)²
which can be easily derived by rewriting (x + y)⁴ as (x + y)²(x + y)² = ((x − y)² + 4xy))(x + y)².
Now let
x + y = s
and substitute this as well as x² − y² = 40 and xy = 99 in the identity above, then we have
s⁴ = 1600 + 396s²
s⁴ − 396s² − 1600 = 0
s⁴ − 400s² + 4s² − 1600 = 0
s²(s² − 400) + 4(s² − 400) = 0
(s² − 400)(s² + 4) = 0
s² − 400 = 0 ⋁ s² + 4 = 0
s² = 400 ⋁ s² = −4
s = 20 ⋁ s= −20 ⋁ s = 2i ⋁ s = −2i
Indeed a clever way to get the sum s w/o calculating x and y first if you know this trick.
However, in a test situation with time restraints I would prefer a bit clumsy but easier to spot substitution y = 99 / x or y² = 9981 / x² which gives us (x² - 81) (x² +121) = 0.
@@MrGeorge1896 Both are nice solutions! I was even lazier, and calculated d = x - y:
(y+d)^2 - y^2 = 40
(y+d) * y = 99
Using the first equation:
y = (40-d^2)/2d = 20/d - d/2
x = y + d
Putting that into the second eqation:
(20/d-d/2 + d)(20/d-d/2) = (20/d+d/2)(20/d-d/2) = 10^2/(d/2)^2 - (d/2)^2 = 99
Let m = (d/2)^2: 100/m - m = 99 or
m^2 + 99 m - 100 = 0
m_1/2 = -99/2 +/- sqrt(9801/4+100) = -99/2 +/- sqrt(10201/4) = -99/2 +/- 101/2 = { -100, 1 }
m = (d/2)^2 > 0, thus m = 1, d = 2.
We already suggested, that 99 = 11 * 9 was one of the sensible integer solution, x = 11, y = 9,
but we have to acknowledge, that x = -11, y = -9 is also a solution (from the structure of the problem).
The funny thing is, that x = 9i, y = -11i, and their negative counterpart also work!
(9i)^2 - (-11i)^2 = -81 - (-121) = 40, (9i)(-11i) = - (-99) = 99, and
(-9i)^2 - (11i)^2 = -81 - (-121) = 40, (-9i)(11i) = - (-99) = 99.
Well, I have to admit, it's not surprising, as you can consider this as a polynomial equation of order 4.
Oops, where are my thoughts gone? m = -100 is exactly about these complex solutions!
m = (d/2)^2 = -100 => d = 20i = 9i - (-11i) = 11i - (-9i).
@@rainerzufall42 If you are going to work with the difference of x and y as a variable, then it makes sense to work with the sum of x and y as a variable as well. The given system of equations is
x² − y² = 40
xy = 99
Using the identities a² − b² = (a + b)(a − b) and (a + b)² − (a − b)² = 4ab this system can be written as
(x + y)(x − y) = 40
(x + y)² − (x − y)² = 396
Now, if we let
x + y = s
x − y = d
we get
sd = 40
s² − d² = 396
Observe that the structure of this system is similar to that of the original system, except that the variables are now s and d. You can solve this system for s in any way you like. A conventional approach would be to substitute d = 40/s from the first equation into the second equation, then we have
s² − (40/s)² = 396
which gives
s⁴ − 396s² − 1600 = 0
Note that this is the exact same biquadratic equation in s which I obtained directly by using the identity (x + y)⁴ = (x² − y²)² + 4xy(x + y)² with the original system in x and y.
😢😊
20
x²-y²=40
xy=99
x+y?
(x²-y²)²=40²
x⁴+y⁴-2(xy)²=1600
x⁴+y⁴-19602=1600
x⁴+y⁴=21202
(x²+y²)²-2(xy)²=21202
(x²+y²)²-19602=21202
(x²+y²)²=40804
|x²+y²|=202
x²+y²=-202
(x+y)²-2xy=-202
(x+y)²-198=-202
(x+y)²=-4
|x+y|=2i
x+y=±2i ❤❤
x²+y²=202
(x+y)²-2xy=202
(x+y)²-198=202
(x+y)²=400
|x+y|=20
x+y=±20 ❤❤