Question 971000: 1. Show that 2 + (the square root) of 2i and 2-(the square root) of 2i satisfy the equation x^2 - 4x + 6 = 0.
2. Solve and check: (please be clear)
a) x^2 +4 = 0
b) x^2 +12 = 0
c) x^2 -2x + 2 = 0
Thank you.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! x^2 -4 x +6=0
The roots are 2 +/- i (sqrt 2), which is different from the sqrt (2i) which doesn't work and isn't found by the quadratic formula (see below).
(2 + i sqrt (2)) ^2 -4 (2 +i sqrt (2)) +6=
4 + 4i^1/2 + 2i^2 -8 -4i ^1/2 +6= The 4i (sqrt (2)) disappear, and i^2=-1
4+ (-2) -8+6=0
2- i (sqrt 2) does the same thing only the signs change in the first term and in the second term.
quadratic formula
a=1 b=-4 c=6
x= [4 +/- sqrt (16-24)]/2]
x={4 +/- sqrt (-8)}/2} = {4 +/- sqrt (-4)*(2)]/2} = 2 +/- (2 i *sqrt (2), after dividing all by 2.
x= 2 +/- i sqrt(2)
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x^2=-4 ; x= +/- 2i ; (2i)^2 is 4 i^2=-4 ; i^2=-1 ; For minus 2i, the square removes the negative.
x^2=-12; x= +/- sqrt (-4)* sqrt (3)= +/- 2i sqrt (3). This squares to 4i^2 *3=-12
x^2-2x-2
(1/2) {2 +/- sqrt (4-8)} = (1/2) 2 +/- sqrt (-4) = (1/2) 2 +/- 2i = 1+/- i
(1+ i )^2 -2(1+i) +2
=1 +2 i + i^2 -2 -2i +2
=1 +2i -1 -2 -21 +2
=0
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