moi à l'iut je faisais de tete lorsque on a a + b = S ab = P a et b sont les solutions de l'equation t² - St + P Delta = S² - 4P x,y =( S +/- racine (Delta) ) / 2 x = e^+/-ipi/3 en 4eme !!!! | x².y + x.y² = -540 | x.y = -45
The square of (-1) is (1), because ((-1) \times (-1) = 1). However, when we talk about (i), the imaginary unit, the situation is different. By definition, (i) is such that (i^2 = -1). By definition i^2 = −1
Is the D=(b^2)-4ac thing just to find out if the answer is a complex answer? Cause I was just taught to use the Quadratic Formula, which is what the x=[-b±SQRT(D)]/2a is.
The simple approach. Substitute x=1/2+z and y=1/2-z into the second equation: (1/2+z)(1/2-z)=1 or z^2=-3/4. Thus, z=±i√3/2, x=(1±i√3)/2, and y=(1∓i√3)/2
This dude’s singsong delivery has me LMAO
You can using complex analysis method sir.
You can use*
not using
@@Mustafa_ShahzadI was about to write the same, lol
y = 1-x;
insert in 2nd equation;
x(1-x)=1 => x² - x + 1 = 0 ;
standard formula yields x1,2 = 0.5 +/- 0.5 * SQRT(-3) = 0.5 +/- 0.5*i*SQRT(3)
=> y1,2 = 0,5 -/+ 0,5*I*SQRT(3)
moi à l'iut je faisais de tete
lorsque on a
a + b = S
ab = P
a et b sont les solutions de l'equation
t² - St + P
Delta = S² - 4P
x,y =( S +/- racine (Delta) ) / 2
x = e^+/-ipi/3
en 4eme !!!!
| x².y + x.y² = -540
| x.y = -45
Yeh I had solve it correctly
The square of (-1) is (1), because ((-1) \times (-1) = 1). However, when we talk about (i), the imaginary unit, the situation is different. By definition, (i) is such that (i^2 = -1). By definition i^2 = −1
Very nice problem! ❤❤
Better way to find y:
Xy=1--> y=1/x--> y is the conjugate of x
Is the D=(b^2)-4ac thing just to find out if the answer is a complex answer? Cause I was just taught to use the Quadratic Formula, which is what the x=[-b±SQRT(D)]/2a is.
in boolean algebra that would work just fine if x=y=1
(0.5)+(0.5)=1 (y ➖ 0.5x+0.5). (xy ➖ 1xy+1).
.
The simple approach. Substitute x=1/2+z and y=1/2-z into the second equation: (1/2+z)(1/2-z)=1 or z^2=-3/4. Thus, z=±i√3/2, x=(1±i√3)/2, and y=(1∓i√3)/2