ax^2+6x-8 What are the all possible combinations of real and complex roots?

Question

how to find all the possible combinations? I know they must equal to 6

Answer 1

If a < 0 then there are two sign changes, so there are two, or no, Real positive roots; if discriminant b²-4ac is positive, those are Real roots, otherwise a complex conjugate pair.  There can be no negative Real roots if a < 0.

 

If a > 0, then there is only one sign change, so there is one real positive zero and therefore one Real negative zero.

 

Forget about "they" (vague antecedent) being 6.

 

 

Answer 2

ax²+6x-8 is a quadratic.  There will be exactly two roots.  They will both be either real or complex.

 

Using the quadratic formula:

 

x = [-b±√(b²-4ac)] / 2a

x = {-6±√[36-(4)a(-8)]} / 2a

x= [-6±√(36+32a)}/2a

 

The part under the radical is the discriminant: (36+32a).  It tells you whether the roots are real or complex.

if (36+32a) is less than zero, the square root of that negative quantity gives the solution an imaginary part.

The discriminant will be negative when 36 + 32a < 0,

32a < -36

a < -36/32

a < -9/8

 

When a is less than -9/8, the roots will be complex numbers.

When a is greater than or equal to -9/8, the roots will be real numbers.

 

Since the -b (-6) is outside the radical and is not zero, there will always be a real part in the solution.  That is, there cannot be a pure imaginary number solution.

 

I don't know where you got that "must equal 6"!