Find all values for prime $p$, so that $x^2+px-330p=0$ only has one positive integral root.

Question

After studying the problem, I found that the sum of the root is $-p$ and the product of them is $-330p$. Does this help?

Answer 1

Hints:

The quadratic $\;x^2+px-330p\;$ has one single (real) root iff its discriminant is zero:

$$\Delta:=b^2-4ac=p^2+1320p=0\iff\;\ldots$$

Now what's the condition(s) that make(s) sure the root is positive?

Another way:

The above quadratic has one single root iff it is a perfect square, so let us complete the square:

$$x^2+px-330p=\left(x+\frac p2\right)^2-\frac{p^2}4-330p\;\implies\;\text{it has to be that}$$

$$\frac{p^2}4+330p=0\iff p^2+1320p=0$$

and we get the same as above...

Answer 2

you need to solve the equation $b^2-4ac=0$, $\Rightarrow p^2+1320p=0$, so $p=0$ or $-1320$, then the root would be $-b=-p=1320$ which is positive.

We reject $p=0$ as it does not give us a positive root.

But then $p$ is not prime..

Answer 3

The sum of the roots is $-p$. Suppose the positive root is $a$, then the negative root is $-p-a$

The product of the roots is $-a(p+a)=-330p$

Now either $p|a$ or $p|(p+a)$ (because $p$ is prime) which also implies $p|a$ so let $a=pb$ then $$pb(pb+p)=330p$$ so that $$pb(b+1)=330$$

And you can go from there ...