(8x^2+ax+38) + (x^2-x-2)i = 0
If this is zero for some values of x, then x^2-x-2 = 0, so x = 2 or -1
If x=2,
32+2a+38 = 0
70+2a=0
a = -35
If x = -1,
8-a+38 = 0
46-a = 0
a = 46
Unfortunately, in both cases, one of the roots is complex. I don't see any value of a where both roots are real. Any other ideas?
If the quadratic equation
(8+i)x^2+(a-i)x+38-2i=0
has real roots, what is the value of the positive real constant a?
1 answer
1 answer