1
$\begingroup$

Find the least number $a$ such that $x^4+ax^3+2017x^2-360x+16 \ge0$ for all positive $x$.

I tried to solve this by differentiating but it is too complicated. Is there good way to solve?

  • 0
    This seems complicated as is ... may I do what you know so far? In terms of mathematics ? (calc 1, calc 2 ..? – 2017-01-12
  • 0
    i tried to find a local min and local maximum but it is to difficult. – 2017-01-12
  • 0
    @user122794 -- It's unethical (and you surely know it!) to post problems from an online entrance exam where the cutoff date for submission is January 20, 2017. – 2017-01-12

1 Answers 1

1

Your inequality can be written:

$f(x)=(x^2+45x-4)^2+(a-90)x^3\ge 0$

$x^2+45x-4=0$ has two solutions: $x_0\approx -45.089<0$ and $x_1\approx 0.088714>0$

In order to have $f(x_1)\ge 0$ one needs $a\ge 90$

We notice that $a\ge 90$ is also a sufficient condition for $f(x)\ge0$ when $x$ is positive.

So the least value $a$ is $90$

  • 2
    How did you find that rearrangement of the equation? – 2017-01-12
  • 0
    I brute-forced $\min a=90$ numerically. Then I looked for special properties of the polynomial for $a=90$, and everything fell in place. – 2017-01-12
  • 0
    wow thanks alot! – 2017-01-12