# Thread: Problem with Roots. Confusion.

1. ## Problem with Roots. Confusion.

If a,b,c,d, and e are real numbers and a≠0, then the equation of
$
ax^7+bx^5+cx^3+dx+e=0$
has

A. No Real Roots
B. At least One Real Root
C. An Odd number of non-real Roots
D. Only One Real Root
E. No Positive Real Roots.

I understand the basic concept of roots, where x is a number such that f(x)=0, but how would you apply that to this problem?

Thanks.

2. Originally Posted by r2d2
If a,b,c,d, and e are real numbers and a≠0, then the equation of
$
ax^7+bx^5+cx^3+dx+e=0$
has

A. No Real Roots
B. At least One Real Root
C. An Odd number of non-real Roots
D. Only One Real Root
E. No Positive Real Roots.

I understand the basic concept of roots, where x is a number such that f(x)=0, but how would you apply that to this problem?

Thanks.
Complex roots of polynomials with real coefficients occur in conjugate pairs, so the number of complex roots is even (here complex means non-real).

A polynomial of odd order has at least one real root.

That should allow you to decide on A, B and C. A suitable choice of signs for a, b, c, d, e and Descartes rule of signs will allow you to decide on E.

Which leaves D (which is false can you see why?).

CB

3. Originally Posted by r2d2
If a,b,c,d, and e are real numbers and a≠0, then the equation of
$
ax^7+bx^5+cx^3+dx+e=0$
has

A. No Real Roots
B. At least One Real Root
C. An Odd number of non-real Roots
D. Only One Real Root
E. No Positive Real Roots.

I understand the basic concept of roots, where x is a number such that f(x)=0, but how would you apply that to this problem?

Thanks.

Well, as an odd degree real polynomial the equation has at least one real root...an beyond that nothing can be said, imo, without knowing something further about the coefficients, since for example the equation x^7 - 1 = 0 has one unique real root, namely 1.

By the way, it is possible to find counterexample for all the other cases.

Tonio

4. Thank you. I understand much better now.