Question

Given that a and b are real numbers such that the follwoing quadratic equation x^2+(a+6i)x+b+12i=0 has a repeated root, what is the value of a−b?

Answer 1

Follow the same strategy as the previous problem, we make use of given information and deduce values by expansion.

Let
f(x)=x^2+(a+6i)x+b+12i=0
and given that f(x) has a repeated root,
we deduce that the discriminant is zero, i.e.
(a+6i)²-48i=0
which expands to
a²-4b-36 + (12a-48)i = 0
Equating the real and complex parts to zero yields two equations:
12a-48=0 (since i≠0)
=>
a=4
Substitute into the equation of the real part:
a²-4b-36 = 0
4²-4b-36 = 0
=>
-4b-20 = 0
=>
b=-5
So a=4, b=-5
check: substitute a=4, b=-5 into
f(x)=x^2+(a+6i)x+b+12i=0