math Q

Based on principle of roots of quadratic equation,
sum of roots = a + b = -(-2)/1 = 2............(1)
product of roots = ab = 6/1 = 6..............(2)
(a-b)^2 = a^2 + b^2 - 2ab
= a^2 + 2ab + b^2 - 2ab - 2ab
= (a + b)^2 - 2ab - 2ab
= (a +b)^2 - 4ab
= (2)^2 - 4(6)
= 4 - 24 = -20.

discriminant of the equation=b^2-4ac=4-4(6)(1)=-20<0
therefore the equation has 2 unequal complex roots
用formula:
a,b=(2 √-20)/2 & (2-√-20)/2
=1 √-5 & 1-√-5
(a-b)^2=[(1 √-5)-(1-√-5)]^2
=(1 √-5-1 √-5)^2
=(2√-5)^2
=(4)(-5)
=-20

*a同b都係complex number,所以佢地互減ge square可以係負數

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