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The question is :

If $a,b$ are integers such that all the roots of the equation $ ( x^2+ax+20)(x^2+17x+b)=0 $ are negative integers. What are the smallest possible vaues of $a,b$?

My approach to this goes like this:

Either $( x^2+ax+20)=0$ $\rightarrow$ $(1)$

or $(x^2+17x+b)=0 $ $\rightarrow$ $(2)$

Let the roots of $(1)$ be $\alpha $ , $\beta $

so, $\alpha $ + $\beta $ = $- a $

and $\alpha \beta $ = $20$

Now, I just put in possible values and subsequently got $-4,-5$ as the numbers which lead to the smallest possible sum, $i.e $ $ 9$.

Let the roots of $(2)$ be $\gamma $ , $\delta $

so, $\gamma $ + $\delta $ = $- a $

and $\gamma \delta $ = $20$

Again, I just put in the possible values and got $-1,-16$ as the numbers which lead to the smallest possible product, $i.e $ $ 16$.

Of course the value of $a+b=25$ but I want to know the proper mathematical method to reach the answer. Please help me with the same. Thanks!

2 Answers2

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Minimize $a=(-\alpha)+(-\beta)$ subject to $(-\alpha)(-\beta)=20$ using $a^2=80+((-\alpha)-(-\beta))^2$, implying $-\alpha,\,-\beta$ need to be as close together as possible, namely $4,\,5$ in some order.

Minimize $b=(-\gamma)(-\delta)$ subject to $(-\gamma)+(-\delta)=17$ using $4b=17^2-((-\gamma)-(-\delta))^2$, implying $-\gamma,\,-\delta$ need to be as far apart as possible, namely $1,\,16$ in some order.

J.G.
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$$x^2+a x+20=0\to x=\frac{-a\pm\sqrt{a^2-80}}{2}$$ as $a$ is the sum of the solutions, if we want them both negative we must have

$a>0;\;a^2-80\ge 0$ that is $a\ge 4\sqrt 5$.

As is given $a$ integer, then must be $a\ge 9$. If $a=9$ then $x_1=-4;\;x_2=-5$.

In a similar way

$$x^2+17 x+b=0$$ has both solutions negative if $b\ge 1$. If $b=16$ then $x_1=-1;\;x_2=-16$

Thus the minimum values are $a=9;\;b=16$

Raffaele
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