Two cubic equations
Problem
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 ܿ$c$.
There are some hints in the Getting Started section.
The Test of Mathematics for University Admission (or TMUA) is designed to give you the opportunity to demonstrate that you have the essential mathematical thinking and reasoning skills needed for a demanding undergraduate Mathematics or Mathematics-related course. There are several UK universities which encourage their applicants to sit TMUA, and a good performance may result in a reduced offer.
For more information about TMUA visit www.admissionstesting.org.
Getting Started
This question comes from a TMUA paper, which means that in the real exam you would have 3-4 mins to do each question. This suggests that there might be a quick way of solving this problem.
Some things to think about:
- What do I know?
- What am I trying to find out?
- Can I rewrite the given information in a way which is more helpful to me? (It could be that different notation might help)
- Will another way of representing the information help? (this could be a diagram/sketch, but could also involve other respresentations).
Student Solutions
Well done to Alfie from England who submitted this solution:
Alfie also explained algebraically why the answer is B:
The solutions of $f(-x)=0$ are the negated solutions of $f(x)=0$ (since if $a$ is a solution of $f(x)=0,$ then $f(a)=0$ and so $f(-(-a))=0$ which means $-a$ is a solution of $f(-x)=0$).