Find conditions on a,b,c so that p(x) and q(x) have exactly 2 roots in common. Also solve the equation p(x)=0

Question

Let $p(x) = x^4 +ax^3 +bx^2+cx +1$
and $q(x) = x^4 +cx^3 +bx^2+ax+1$
with a,b,c real numbers.Find conditions on a,b,c so that p(x) and q(x) have exactly 2 roots in common. Also solve the equation p(x)=0

I get that a+b+c=-2. HOw to solve it

Answer 1

Let $\alpha, \beta$ be the common roots, so we have $p(\alpha) = p(\beta) = q(\alpha) = q(\beta) = 0$. Thus:

$\alpha^4 +a\alpha^3 +b\alpha^2+c\alpha +1 = 0 \tag{1}$
$\beta^4 +a\beta^3 +b\beta^2+c\beta +1 = 0 \tag{2}$
$\alpha^4 +c\alpha^3 +b\alpha^2+a\alpha +1 = 0 \tag{3}$
$\beta^4 +c\beta^3 +b\beta^2+a\beta +1 = 0 \tag{4}$

$(1) - (3)$ gives $(a-c)\alpha^3 + (c-a)\alpha = 0 \implies \alpha(a-c)(\alpha^2-1)=0$
$(2) - (4)$ gives $(a-c)\beta^3 + (c-a)\beta = 0 \implies \beta(a-c)(\beta^2-1)=0$

Now $\alpha = 0$ would not satisfy $(1)$ and $a= c$ would mean $p = q$ so all four roots will be equal, so we must have $\alpha^2 = \beta^2 = 1$. So we have choices of $(\alpha, \beta) \in \{(-1, 1), (1, -1)\}$, if only distinct roots are allowed. WLOG let $(\alpha, \beta) = (1, -1)$

$(1) \implies a+b + c + 2 = 0$,
$(2) \implies -a+b - c + 2 = 0$
Adding, we have $b = -2$, so $a = -c$.

So $p(x) = x^4 + ax^3 - 2x^2 - ax + 1, ~~ q(x) = x^4 - ax^3 - 2x^2 + ax + 1$, expressed in only one parameter $a$. The conditions are all satisfied with any non-zero value for $a$.

Further, if the other roots of $p(x)$ are $p_1, p_2$, we must have $p_1 p_2 = -1$ due to the constant term. Similarly $p_1+p_2 = -a$ So these must be the roots of the quadratic $x^2 + ax - 1 = 0$ or $p_1, p_2 = \dfrac{-a \pm \sqrt{a^2+4}}{2}$, expressed in terms of the same parameter $a$.