4

Suppose that $f(x) = x^3 + ax^2 + bx + c$ has three distinct integral roots and $f(x^2+2x+2)$ has no real roots.

  1. What is the minimum value of $a$?
  2. What is the minimum value of $b$?
  3. What is the minimum value of $c$?
  4. In the case when $a$, $b$ and $c$ take their minimum values, if the roots of $f'(x) = K$ are equal, then what is $K$?

My attempt:

I factorised $g(x) = x^2+2x+2$ as $(x+1)^2 + 1$. For all values of $g(x)$, $f(x)$ will always give a non-zero value. I tried getting multiple linear equations in $a$, $b$ and, $c$, but it got me nowhere. I don't see any other way to obtain minimum values of said variables, let alone attempt question 4.

This is a problem from a revision worksheet for our polynomials class.

Currently, I have no other ideas as to how to approach solving this problem. Any help would be appreciated.

Kenny Wong
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dayDreams26
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User May 23 '23 at 06:59
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    Got it. Thanks! – dayDreams26 May 23 '23 at 07:09
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    $f(x^2+2x+2) = 0$ if and only if $x^2+2x+2$ equals one of the roots of $f(y) = 0$, but all these are integers. Thus, you want a list of integer values of $y$ such that $x^2+2x+2 = y$ has nonreal roots (which is easily found by checking the sign of the discriminant). Once you have all those integer values of $y$, then only those values can be roots of $f$. Finding all such $y$ is then a starting point for your question. Then, you use Vieta's formulas to relate the values of $a,b,c$ to these roots. That answers the first three parts of the question. – Sarvesh Ravichandran Iyer May 23 '23 at 07:18
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    Hint: What are all the possible values of $x^2+2x+2$, if $x$ can be any real number? – aschepler May 23 '23 at 07:19

2 Answers2

1

$f(y) = 0$ has three distinct integral solutions for $y$

Thus, for $f(x^2 + 2x + 2)$ to not have integral solutions :
$x^2 + 2x + 2 = y$ cannot have any real solutions. Checking the sign of discriminant we get
$y = [0, -1, -2 ...]$ which is essentially [ $y < 1$ ]

This implies that for minimum values of coefficients; $a$, $b$, and $c$. We use Vieta Relations and get the answers to the questions as :
$a = 3$
$b = 2$
$c = 0$
$k = -1$

Credits to Sarvesh and aschepler for their much needed help.

dayDreams26
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0

Your factorization shows that $g$ is surjective from $\mathbb{R}$ to $[1,+\infty[$. Therefore, $f(g(x))$ has no real root is equivalent to all the roots of $f$ are $< 1$ thus $\leqslant 0$ because they are integral. Let $r_1,r_2,r_3$ be the roots of $f$. Newton relations give you, $$ r_1 + r_2 + r_3 = -a, \qquad r_1r_2 + r_2r_3 + r_3r_1 = b, \qquad r_1r_2r_3 = -c. $$ Knowing that for all $i$, $r_i \leqslant 0$, you can deduce bounds on $a,b,c$, like for example $a = -r_1 - r_2 - r_3 \geqslant 0$. As the $r_i$ are distinct you even have $a \geqslant 0 + 1 + 2 = 3$. Try to do the same for $b$ and $c$ to decue bounds $a_0$, $b_0$ and $c_0$. Verify that those bounds are optimal by checking that $x^3 + a_0x^2 + b_0x + c_0$ indeed have three disctinct integral non-positive roots, and question $4$ is easy by computing the discriminant of $f' - K$.

Cactus
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