If $x^3+ax^2=bx+c$ (with $a,b,c>0$) has one positive root and two negative roots, then what can you say about the roots of $x^3+c=ax^2+bx$?

Question

I can't manage to solve this maths problem. I was hoping someone could answer it and explain how they did it to me. Bear in mind I have only just finished A levels so I wont understand any crazy business/shenanigans any of you maths graduates might get up to. However, I don't think any of that will be necessary to answer this question.

The positive real numbers a, b and c are such that the equation

$$x^3 + ax^2 = bx + c$$

has three real roots, one positive and two negative.

Which one of the following correctly describes the real roots of the equation

$$x^3 + c = ax^2 + bx$$

A. It has three real roots, one positive and two negative.

B. It has three real roots, two positive and one negative.

C. It has three real roots, but their signs differ depending on $a$, $b$, and $c$.

D. It has exactly one real root, which is positive.

E. It has exactly one real root, which is negative.

F. It has exactly one real root, whose sign differs depending on $a$, $b$, and $c$.

G. The number of real roots can be one or three, but the number of roots differs depending on $a$, $b$, and ܿ$c$.

You can find the question written more nicely in this PDF (Question 19):

Answer 1

Try to use the Rule of Signs.

It says that:

The number of sign changes is the maximum number of positive roots

Your 2nd equation, after re-ordering the powers of $x$ can be written as:

$$+x^3 -a x^2 - bx - c$$

The sign changes between the first 2 terms is 1 (changed from + to - one time).

The sign change between the 2nd term and the 3rd term is zero (we have -a, -b).

The sign change between the 3rd term and the 4th term is zero (we have -b, -c).

Total number of sign changes is: $1+0+0=1$.

This tells us that this polynomial has at most $1$ POSITIVE real root (EDITED).

This makes choices A, C, D, E ,F False.

G is false because a polynomial of degree $n$ will always have $n$ roots and does not depend on $a,b,c$.

To make that sure that the real root is negative using the same rule, change every $(x)$ to $(-x)$ .

$$f(-x)=-1.x^3+ax^2+bx-c$$

Now counting the number of sign changes we get:

From (-1 to +a) = 1 change in sign

From (+a to +b) = 0 change in sign

From (+b to -c)= 1 change in sign

Total sign changes for $f(-x)=1+0+1=2$

This indicates that we have at most 2 NEGATIVE real roots (or zero negative roots if these 2 remaining roots are complex, since complex roots always come in pairs).

To summarize

We either have 2 negative roots or 0 negative roots AND we have at most 1 real positive roots.

This leaves B as the answer closest to correctness as you have noticed.

Answer 2

Odd degree polynomials with real coefficients have at least one real root. Complex solutions come in conjugate pairs. Use those theorems.