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3 Distinct Roots Not (Posted on 2008-03-04) Difficulty: 3 of 5
Prove that the equation

   x3 + 2px2 + 2p2x + p = 0

cannot have three distinct real roots, for any real number p.

See The Solution Submitted by Bractals    
Rating: 4.0000 (2 votes)

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Solution 3 distinct solutions Comment 4 of 4 |
A necessary and sufficient condition for a cubic polynomial with real coefficients to have 3 distinct real roots is that its discriminant is positive: if the roots are u,v,w, the discriminant is ((u-v)(u-w)(v-w))^2, which is clearly positive if the roots are distinct real numbers. On the other hand, if u is real but v,w are conjugate complex roots, then the discriminant is -4|u-v|^4 Im(v)^2 < 0.

In this case the discriminant is -p^2 (27-40p^2+16p^4) < 0 for all real p.

Consider a cubic f(x) = x^3 + ax^2 + bx + c.

If this has 3 distinct real roots, then Charlie's criterion is that f'(x) must have 2 real roots, and therefore it is necessary that a^2 - 3b > 0. However, this can't be sufficient in general, because for any a,b, we can always pick c such that f(x) has only one real root.

Similarly, Chesca's criterion is that the sum of the squares of the roots must be positive, or a^2 - 2b > 0. This condition is weaker than Charlie's, because if a^2 - 3b > 0, then a^2 - 2b > 0. However, there's a way to strengthen this condition to make it equivalent to Charlie's: we require that the sum of the squares of the roots of f(x-t) also be positive, for all t real t. If the roots of f(x) are u,v,w, the sum of the squares of the roots of f(x-t) is

(u+t)^2 + (v+t)^2 + (w+t)^2 = u^2+v^2+w^2 + 2t(u+v+w) + 3t^2 = (a^2-2b) - 2at + 3t^2,

and this is positive for all t iff its discriminant (2a)^2 - 4*3(a^2-2b) = -8(a^2 - 3b) < 0. So the condition that the sum of the squares of the roots of every translate of f be positive is equivalent to a^2 - 3b > 0, which is equivalent to f'(x) having two distinct real roots.

So to summarize, if a^2 - 3b <= 0, then f can't have 3 distinct real roots, no matter the value of c. However, if a^2 - 3b > 0, there will be an open interval of values of c for which f has 3 distinct real roots, namely

(27c + 2a^3 - 9ab)^2 < 4(a^2 - 3b)^3,

which is equivalent to the discriminant being positive. Another way of deriving the above is as follows: if x1 < x2 are the roots of f'(x)=0 (which exist iff a^2-3b>0, and depend only on a,b), then f has 3 distinct real roots iff f(x1) > 0 > f(x2), which is equivalent to the above condition on c.
  Posted by Eigenray on 2008-03-07 19:22:53
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