I got a binomial question and when trying to solve for p and q I reached $x^8$. Am I wrong?

Question

• $\left(px-\frac qx\right)^8 $
• fifth term is $5670 $
• $p-q=2 $
• $p$ and $q$ are positive

using this info i got:
$$x^{8}+560x^{7}+1680x^{6}+2240x^{5}+1120x^{4}-5670=0 $$ i am unable to solve this. help. thanks

Answer 1

Assuming by "fifth term" you mean the fifth in order of descending degree of $x$, then clearly since there are $8 + 1 = 9$ terms in the expansion, the fifth term is the constant term. Hence $$\binom{8}{4} p^4 q^4 = 5670,$$ or $pq = 3$. Then since $p = q + 2$, it follows that $$q(q+2) = 3$$ or $$q \in \{-3, 1\}.$$ Since $q > 0$ we conclude $q = 1$ and $p = 3$, hence $$(px - q/x)^8 = (3x - 1/x)^8.$$ Consequently if we require the roots for this equation, then $3x - 1/x = 0$, or $x = \pm 1/\sqrt{3}$. There are no other roots; each one has multiplicity $4$.