I can't wrap my head around it as no roots are given.

Question

If $x^4+2x^3+px^2+qx+9=0$ is a complete square. $p$ and $q$ are positive. Find the value of $p$ and $q$.

Answer 1

If it is a complete square, then

$$x^4+2x^3+px^2+qx+9=(x^2+ax+b)^2=x^4+2ax^3+(2b+a^2)x^2+2abx+b^2$$

(we have the coefficient in front of $x^2$ is $1$ since the coefficient in front of $x^4$ is $1$). Clearly, $b=3$ and $a=1$ which gives us

$$x^4+2x^3+px^2+qx+9=(x^2+x+3)^2=x^4+2x^3+7x^2+6x+9$$

$$px^2+qx=7x^2+6x$$

So take $p=7$ and $6=q$.

Answer 2

Okay, I'm assuming you mean that the polynomial $f(x) = x^4 + 2x^3 + px^2 + qx + 9$ is a square, that is there is some other polynomial $g(x)$ with $g(x) * g(x) = f(x)$.

So what could this polynomial $g(x)$ be? Well, the highest-degree term in $f(x)$ is $x^4$, so that means the highest-degree term in $g(x)$ should be $x^2$ ($x^2 * x^2 = x^4$).

So $g(x) = x^2 + ax + b$, for some constants $a$ and $b$. Now lets look at what happens when we multiply it out. We get $f(x) = g(x) * g(x) = x^4 + 2ax^3 + (2b + a^2) x^2 + 2abx + b^2$. This tells us a few things. First of all, by looking at the $x^3$ term we see $2a = 2$, or $a = 1$. Second, by looking at the constant term, we see $b^2 = 9$, so $b = \pm 3$. So that gives us two possibilities for $g(x)$. $g(x) = x^2 + x + 3$ or $g(x) = x^2 + x - 3$.

Now lets look at what happens with $p$ and $q$ in both cases. Remember, by looking at the multiplication we have $p = (2b + a^2)$ and $q = 2ab$. So if $p$ and $q$ are positive, we have to take the $+3$ option. In which case $p = 7$ and $q = 6$.