If #x^2+px+q=0# has root #x=p-q#, then what is the maximum value of #pq# ?

1 Answer
George C.
Jan 8, 2018

#pq# can be arbitrarily large and positive - it has no maximum value.

Explanation:

We have:

#0 = x^2+px+q#

#color(white)(0) = (x-alpha)(x-beta)#

#color(white)(0) = x^2-(alpha+beta)x+alpha beta#

where #alpha#, #beta# are the roots.

Equating coefficients:

#{ (alpha+beta = -p), (alphabeta = q) :}#

Then putting #alpha = p-q#, we find:

#beta = -alpha-p = q-2p#

So:

#q = alphabeta = (p-q)(q-2p) = 3pq-2p^2-q^2#

So:

#2p^2-3qp+(q^2+q) = 0#

For this to have real solutions #p#, we require its discriminant to be non-negative.

So:

#(-3q)^2-4(2)(q^2+q) >= 0#

That is:

#q^2-8q >= 0#

i.e.

#q(q-8) >= 0#

This will be true for #q <= 0# and #q >= 8#

Given some #q#, we have:

#0 = 8(2p^2-3qp+(q^2+q))#

#color(white)(0) = 16p^2-24qp+8(q^2+q)#

#color(white)(0) = (4p)^2-2(4p)(3q)+(3q)^2-(q^2-8q)#

#color(white)(0) = (4p-3q)^2-(sqrt(q^2-8q))^2#

#color(white)(0) = (4p-3q-sqrt(q^2-8q))(4p-3q+sqrt(q^2-8q))#

So:

#4p = 3q+-sqrt(q^2-8q)#

So:

#p = 1/4(3q+-sqrt(q^2-8q))#

This will be real valued for any #q >= 8#.

If we choose the positive sign for the square root, then we have #p > 3/4q#. So #p > 6# and by making #q# arbitrarily large and positive, we can make #pq# arbitrarily large and positive.

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