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Пікірлер: 9
This really shouts: Use the pq-formula. When a=1, and b is even the pq formula is much simpler.
When you realize X2 is the same as Y1, calculating Y2 is redundant. It's obviously going to be X1.
t²-6t+36=0 t (solutions of a,b and b,a)
The problem is incomplete until you state the domain of X and Y. By inspection, X = 3 + ai, Y = 3 - ai where a^2 = 36 - 9 = 27, i.e. a = 3sqrt(3)
Solution of equation ! X2 - SX + P = 0
Here is an easy method for solving this system of equations: x+y=6 -----> [1] x y=36 -----> [2] (x-y)^2=(x+y)^2 - 4 x y (x-y)^2=6^2 - 4 * 36 (x-y)^2=36 - 4 * 36 (x-y)^2= - 3 * 36 x-y = +/- 6 i √3 -----> [3] 2 x = 6 +/- 6 i √3 From[1] + [3] x = 3 +/- 3 i √3 2 y = 6 -/+ 6 i √3 From[1] - [3] y = 3 -/+ 3 i √3 (x, y) = { (3 + 3 i √3, 3 - 3 i √3), (3 - 3 i √3, 3 + 3 i √3) }
@dmitripogosian5084
28 күн бұрын
Why bother ? Just solve for x the first equation, substitute into the second and solve the resulting quadratic equation. And check that you did not miss anything, so that there is second pair of solutions since problem is symmetric relative to x and y
In reality, there is NO solution. I could solve this, but the square root of a negative actually doesn't exist. Just because a number with power 2 can never be positive. But I suppose that's why it's called a complex number.. I will never understand why they invented i = -1
@johnchestnut5340
27 күн бұрын
History. It's easy to look up. It's neat. Too bad I don't remember it.