( x + 3 i )² = x² + 2 ∙ x ∙ 3 i + ( 3 i )² =
x² + 6 x i + ( 3)² ∙ ( i )² =
x² + 6 x i + 9 ∙ ( - 1 ) =
x² + 6 x i - 9
( x + 3 i )² = 2 y i
x² + 6 x i - 9 = 2 y i
For the real parts:
x² - 9 = 0
x² = ±√ 9
x = ± 3
For the imaginary parts:
6 x i = 2 y i Divide both sidesd by 2 i
3 x = y
y = 3 x
Now:
x = - 3
y = 3 x = 3 ∙ ( - 3 ) = - 9
x = 3 x
y = 3 ∙ 3 = 9
Solutions:
x = - 3 , y = - 9
and
x = 3 , y = 9
(x+3i)²=2yi
Complex equation
Answer plz
5 answers
Remark:
For the real parts:
x² - 9 = 0
becouse:
x² + 6 x i - 9 = 2 y i
you can write as:
x² + 6 x i - 9 = 0 + 2 y i
So the real parts is:
x² - 9 = 0
For the real parts:
x² - 9 = 0
becouse:
x² + 6 x i - 9 = 2 y i
you can write as:
x² + 6 x i - 9 = 0 + 2 y i
So the real parts is:
x² - 9 = 0
(x+3i)²=2yi
If a,b are the roots of pt²+qt+q=0prove that
We need to prove that a+b = -q/p.
By using the quadratic formula, we know that the roots of the equation pt² + qt + r = 0 are given by:
t = (-q ± √(q² - 4pr)) / 2p
Substituting t = a and t = b, we get:
a = (-q ± √(q² - 4pr)) / 2p
b = (-q ∓ √(q² - 4pr)) / 2p
Note that a and b have opposite signs in front of the square root because they are different roots.
Adding these two equations, we get:
a + b = (-2q) / 2p = -q/p
Therefore, we have proved that a+b = -q/p.
By using the quadratic formula, we know that the roots of the equation pt² + qt + r = 0 are given by:
t = (-q ± √(q² - 4pr)) / 2p
Substituting t = a and t = b, we get:
a = (-q ± √(q² - 4pr)) / 2p
b = (-q ∓ √(q² - 4pr)) / 2p
Note that a and b have opposite signs in front of the square root because they are different roots.
Adding these two equations, we get:
a + b = (-2q) / 2p = -q/p
Therefore, we have proved that a+b = -q/p.
5 answers