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Quadratic Challenging Questions 2/3

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

b2-4ac

=(-2a)2-4(1)(a2+5)

=4a2-4a2-20

=-20<0

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

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