momo1729 wrote: Czech and Slovac) Prove that the equation $x^{2}+p|x| = qx - 1 $ has 4 real solutions iff $p+|q|+2<0$ ($p$ and $q$ are two real parameters). (Assuming "4 real solutions" means "4 distinct real solutions") : So $x^2+(p-q)x+1=0$ and $x\ge 0$ has two distinct real solutions $\iff$ $(p-q)^2>4$ and $p-q\le 0$ $\iff$ $q-p>2$ And $x^2-(p+q)x+1=0$ and $x\le 0$ has two distinct real solutions $\iff$ $(p+q)^2>4$ and $p+q\le 0$ $\iff$ $p+q<-2$ $\iff$ $p+2<q<-p-2$ $\iff$ $|q|<-p-2$ $\iff$ $p+|q|+2<0$ Q.E.D.

Thank you I added the "distinct".