Question

Show that x^3 +px-q with p,q>0, has one real root

Answer 1

Answer:

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Step-by-step explanation:

Let f = x^3 + p x + q

Then f’ = 3 x^2 + p  

Setting f’ = 0  There are three cases:  

p > 0      No solution,  so f is monotone increasing, so just one real root

p = 0  f = x^3 + q     ,    just one real root for the same reason

p < 0    , two solutions x = sqrt(z/3)  and – sqrt(z/3)  where z = -p . z is >0

The local minimum of f will occur at   x = sqrt(z/3)  .  plugging in give  

f = (z/3)^3/2 – z (z/3)^1/2 + q this must be > 0 to get three real roots.

Move q to the other side , then square both sides and rearrange terms to get

z^3 - (2/9) z^3 + z^3 < q^2 or

Or  4 z^3   < 27 q^2

which means 27 q^2 + 4 p^2  <0

 

Answer 2

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Step-by-step explanation:

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