Show that that if $p,q,r,s$ are real numbers and $pr=2(q+s)$, then at least one of the eqns $x^2+px+q=0$ and $x^2+rx+s=0$ has real roots.
My Attempt to the solution
we know to have a real solution d>=0 so either
1) $p^2-4q>=0$
or
2) $r^2-4s>=0$
or both are true.
Rearranging we get $(pr)^2 \geq 16qs$ substituting it in the first eqn we get $16qs\geq4q^2 +4s^2 +8qs$ So we get $0\geq(q-s)^2$ so we get $q=s$. Now what to do..? is there any another way to solve this?
Given: $pr = 2(q+s)$. To prove: either $p^2 - 4q \geq 0$ or $r^2 - 4s \geq 0$. It is best to argue by contradiction -- assume both $p^2 - 4q < 0$ and $r^2 - 4s < 0$. Then upon rearranging and adding the inequalities, $$ \begin{array}{rcl} p^2 + q^2 &<& 4r + 4s \\ \frac{p^2 + q^2}{2} &<& 2(r + s) \end{array} $$ Can you see how to finish from here? If not, post a comment and I'll write more.
Hope this helps!