Show that x2-2ax+a2+5>0 for all real values of a. Hence, or otherwise, solve the inequality (3x2+2x-1)(x2-2ax+a2+5)>0.
Show that the graph is always positive (condition 1) and not touching the x axis (condition 2).
graph>0 means graph will never be =0
if graph ≥ 0, it means that graph touches the x axis
Since b2-4ac<0, x2-2ax+a2+5 will never have real roots and will never touch the x axis (condition 2 met) , the graph is positive because coefficient of x2 is positive (condition 1 met), graph is >0 for all real values of a.
Hence (3x2+2x-1)(x2-2ax+a2+5) = (3x-1)(x+1)(x2-2ax+a2+5)
- factorise 3x2+2x-1 to (3x-1)(x+1) –> 2 roots
- x2-2ax+a2+5 –> 0 roots
- Draw out graph
if graph >0, x<-1 or x> 1/3