If both of the zeros of #ax^2+bx+c# are positive or both negative then what can you say about the signs of #a, b# and #c# ?
2 Answers
Answer:
See explanation...
Explanation:
Given a quadratic polynomial:
#ax^2+bx+c#

If the zeros are both positive then
#a# and#c# have the same sign and#b# has the opposite sign. 
If the zeros are both negative then
#a# ,#b# and#c# have the same sign.
To see these, suppose the zeros are
We have:
#ax^2+bx+c = a(xalpha)(xbeta)#
#color(white)(ax^2+bx+c) = a(x^2(alpha+beta)x+alphabeta)#
#color(white)(ax^2+bx+c) = ax^2(alpha+beta)ax+alphabetaa#
Matching the coefficients, we find:
#{ (b = (alpha+beta)a), (c = alphabetaa) :}#
If
If
Note
The converses of these two results are not generally true.
There are quadratic equations
For example:
#x^2+x+1#
has no real zeros.
It still has two zeros, but they are so called "imaginary" a.k.a. nonreal complex numbers. They do not lie on the real number line, but somewhere else in the complex plane. The real number line forms the
All complex numbers can be expressed in the form
#i^2 = 1#
For example, the zeros of
#1/2+sqrt(3)/2i" "# and#" "1/2sqrt(3)/2i#
It is unfortunate that the term imaginary is used of such numbers  coined when such numbers were poorly understood. They are just as real as Real numbers and better behaved in some respects.
Answer:
Some more explanation...
Explanation:
Attempting to make this a little easier to follow, let us start with the simpler case where the leading coefficient is
#x^2+px+q#
If
Using FOIL to help multiply, we find:
#(xalpha)(xbeta) = overbrace((x * x))^"First"+overbrace(x * (beta))^"Outside"+overbrace((alpha) * x)^"Inside"+overbrace((alpha) *(beta))^"Last"#
#color(white)((xalpha)(xbeta)) = x^2betaxalphax+alphabeta#
#color(white)((xalpha)(xbeta)) = x^2(alpha+beta)x+alphabeta#
Comparing this with the expression

If
#alpha > 0# and#beta > 0# then#(alpha+beta) < 0# , so#p < 0# .
Also#alphabeta > 0# , so#q > 0# .#color(white)(0/0)#
In summary, the pattern of the signs of the coefficients of#x^2+px+q# is#+  +# .#color(white)(0/0)#
For example#x^23x+2 = (x1)(x2)# has zeros#1# and#2# .#color(white)(0/0)# 
If
#alpha < 0# and#beta < 0# then#(alpha+beta) > 0# , so#p > 0# .#color(white)(0/0)#
Also#alphabeta > 0# , so#q > 0# .#color(white)(0/0)#
In summary, the pattern of the signs of the coefficients of#x^2+px+q# is#+ + +# .#color(white)(0/0)#
For example#x^2+3x+2 = (x+1)(x+2)# has zeros#1# and#2# .
The more general case of
We have:
#ax^2+bx+c = a(x^2+b/ax+c/a) = a(xalpha)(xbeta)#

If
#alpha > 0# and#beta > 0# then the coefficients#1# ,#b/a# and#c/a# have signs in the pattern#+  +# .#color(white)(0/0)#
Hence if#a > 0# then the pattern of the signs of the coefficients of#ax^2+bx+c# is also#+  +# .#color(white)(0/0)#
If instead#a < 0# then the pattern of the signed of the coefficients of#ax^2+bx+c# is# + # .#color(white)(0/0)# 
If
#alpha < 0# and#beta < 0# then the coefficients#1# ,#b/a# and#c/a# have signs in the pattern#+ + +# .#color(white)(0/0)#
Hence if#a > 0# then the pattern of the signs of the coefficients of#ax^2+bx+c# is also#+ + +# .#color(white)(0/0)#
If instead#a < 0# then the pattern of the signs of the coefficients of#ax^2+bx+c# is#  # .