If $a,b,p,q$ are non-zero real numbers, then the 2 equations $2a^2x^2-2abx+b^2=0$ and $p^2x^2+2pqx+q^2=0$ have...

Question

If $a,b,p,q$ are non-zero real numbers, then the 2 equations $2a^2x^2-2abx+b^2=0$ and $p^2x^2+2pqx+q^2=0$ have:
(a)no common root
(b)two common roots if $3pq=2ab$
(c)one common root if $2a^2+b^2=p^2+q^2$
(d)two common roots if $3qb=2ap$

Okay so the answer given is 'no common root'. How I approached the question...I used the condition for common root which gives me
$$(b^2p^2-2a^2q^2)^2=(-2abq^2-2b^2pq)(4a^2pq-2abp^2)$$

Now I don't know how to prove that this equation can't be satisfied for any values of $a,b,p,q$.

Answer 1

Hint: the second equation is $\,(px+q)^2=0\,$ then substituting $\,x=-q/p\,$ into the first one gives $\,a^2q^2 + (aq+pb)^2=0\,$.