Questions
If roots of the equation are purely imaginary then
a
(b−b)2+(a+a)(ab+ab)=0
b
(b−b)2+(a−a)2=0
c
(b+b)2−(a−a)2=0
d
None of these
detailed solution
Correct option is A
Let ix(x∈R) be root of z2+az+b=0 than −x2+aix+b=0⇒−x2−a¯ix+b¯=0 Subtracting we get(a+a¯)ix+b−b¯=0⇒ x=−b−b¯i(a+a¯)=i(b−b¯)a+a¯Putting this in (1), we get(b−b¯)2(a+a¯)2−a(b−b¯)a+a¯+b=0⇒(b−b¯)2−a(b−b¯)(a+a¯)+b(a+a¯)2=0⇒(b−b¯)2−(a+a¯){ab−ab¯−ab−a¯b}=0⇒(b−b¯)2+(a+a¯)(ab¯+a¯b)=0Talk to our academic expert!
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