Large root for quadratic polynomial

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Large root for quadratic polynomial

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oVlad

#1 h Feb 16, 2023, 3:20 PM

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What is the largest possible rational root of the equation $ax^2 + bx + c = 0{}$ where $a, b$ and $c{}$ are positive integers that do not exceed $100{}$ ?

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#2 h Feb 16, 2023, 3:49 PM

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If the equation has roots, then by quadratic formula we have
$x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}$ is the larger root. We can easily see that picking smaller $a$ 's gives us a larger root, so $a=1$ . Meaning that the largest rational root is an integer, but now $x_1 < \frac{-b+b}{2} = 0$ and $-1$ is a root of $x^2+2x+1$ , so $-1$ is the largest possible rational root.

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pco

#3 h Feb 16, 2023, 3:58 PM

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TheMathBob wrote:

$x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}$ is the larger root. We can easily see that picking smaller $a$ 's gives us a larger root

Not if the root is negative (and it is since $a,b,c>0$ )

oVlad wrote:

What is the largest possible rational root of the equation $ax^2 + bx + c = 0{}$ where $a, b$ and $c{}$ are positive integers that do not exceed $100{}$ ?

Since rational, we need $b^2-4ac=u^2$ for some integer $b>u\ge 0$ and then :
$a=\frac{b^2-u^2}{4c}$ and root is $r=-\frac{2c}{u+b}$

So we need $b,c$ both odd or both even (in order $a\in\mathbb Z$ ) and $c$ smallest possible and $u$ greatest possible
$u<b$ with same parity implies $u\le b-2$ and smallest $c$ is $1$

Choosing $u=b-2$ and $c=1$ implies $a=b-1$ and $r=-\frac 1{b-1}$
And constraint $a,b,c\le 100$ implies $b=100$

And $\boxed{r=-\frac 1{99}}$ when quadratic is $99x^2+100x+1=0$

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#4 h Mar 28, 2023, 4:27 PM

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Soln

This post has been edited 1 time. Last edited by math_comb01, Mar 28, 2023, 4:27 PM

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