I recently came across two interview questions for admission in B.Math at an university. I gave the two questions a try and want to know if my solutions are correct or not.
Q1: Given that $x^4-4x^3+ax^2+bx+1=0$ has all positive roots and $a,b\in\Bbb R$, prove that all the roots are equal.
My solution: Let $p,q,r,s$ be the four roots of the given equation. Using Vieta's formulas, we have,
$$p+q+r+s=4\quad\textrm{and}\quad pqrs=1$$
Since $p,q,r,s$ are all positive, we have, by the AM-GM inequality,
$$\frac{p+q+r+s}{4}\geq\sqrt[4]{pqrs}=\sqrt[4]{1}=1\implies p+q+r+s\geq 4$$
Since we got $p+q+r+s=4$ using Vieta's formulas and knowing that the equality case in AM-GM inequality holds iff $p=q=r=s$, we conclude that all the roots to the given equation are equal. $_\square$
Q2: Without actually computing anything, find the value of $\dbinom{p+q}2-\dbinom p2-\dbinom q2$.
My solution: Since it's told not to actually compute anything, I suppose that they were asking for a combinatorial proof. I have the following argument:
Suppose we have $p+q$ people in a room with $p$ people in Group 1 and $q$ people in Group 2. Then, $\dbinom{p+q}2$ counts the number of ways we can select two people from the people in the entire room. However, $\dbinom p2$ and $\dbinom q2$ count the number of ways we can select two people from Group 1 and Group 2 respectively.
Now, we can select two people from the entire room by either taking two people from Group 1 or taking two people from Group 1 or taking one person from Group 1 and another from Group 2. These are the only possible methods.
So, the expression we have counts the number of ways we can select one person from Group 1 containing $p$ people and another person from Group 2 containing $q$ people. By the rule of product, we have $pq$ ways to do this and hence the value of the given expression is $pq$. $_\square$
Citing the number of upvotes on Brian M. Scott's comment, I believe it's safe to say your answers are correct. I would just like to officially answer this question to remove it from the Unanswered Questions queue.