Question

The quadratic equation y=2x^2-px+6 has equal roots. Find the possible values of p.

Answer 1

$\Delta=0\\ \Delta=(-p)^2-4\cdot2\cdot6=p^2-48\\ p^2-48=0\\ p^2=48\\ p=\sqrt{48} \vee p=-\sqrt{48}\\ p=4\sqrt3 \vee p=-4\sqrt3$