If #a, b, c# are positive numbers in arithmetic progression and #alpha, beta# are the zeros of #ax^2+bx+c#, then what is the value of #alpha+beta+alphabeta# and what are the possible integer solutions of #ax^2+bx+c=0# ?

1 Answer
George C.
Dec 15, 2017

#alpha+beta+alphabeta = (c-b)/a#

but if #a, b, c# are positive numbers in arithmetic progression then the quadratic has no real zeros.

Explanation:

There seem to be some extraneous and contradictory requirements in the question, but let's observe a few things...

If #alpha# and #beta# are the zeros of #ax^2+bx+c#, then we find:

#ax^2+bx+c = a(x-alpha)(x-beta)#

#color(white)(ax^2+bx+c) = a(x^2-(alpha+beta)x+alphabeta)#

#color(white)(ax^2+bx+c) = ax^2-a(alpha+beta)x+aalphabeta#

Equating coefficients, we find:

#{ (alpha+beta = -b/a), (alphabeta = c/a) :}#

So:

#alpha+beta+alphabeta = -b/a+c/a = (c-b)/a#

What are the possible values for #a, b# and #c# ?

We are told that they are positive numbers in arithmetic progression. So let's put:

#{ (b = a + d), (c = a + 2d) :}#

where #d in (-a, oo)#

The discriminant of our quadratic is:

#Delta = b^2-4ac#

#color(white)(Delta) = (a+d)^2-4a(a+2d)#

#color(white)(Delta) = (a^2+2ad+d^2)-(4a^2+8ad)#

#color(white)(Delta) = -3a^2-6ad#

#color(white)(Delta) = -3a(a+2d)#

#color(white)(Delta) = -3ac < 0#

So the quadratic has no real zeros, let alone integers.

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