Imaginary Discriminant

Level pending

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

-1

1

-9

9

1 solution

Himansu Rathi
Feb 3, 2014

x^{2}+(ax+6i)x+b+12i=0+0i

so 6x+12=0(equating imaginary parts of equation)

gives x=-2

Discriminat of equation=0

(ax+6i)^{2}==4(b+12i)

gives

a^{2}-36=4b

and 12a=48

So a=4

and b=-5

Hence, a-b=4+5=9 Ans