Ta có: 3 x 2 + 2 5 x - 3 3 = - x 2 - 2 3 x +2 5 +1
⇔ 3 x 2 + 2 5 x - 3 3 + x 2 + 2 3 x - 2 5 – 1= 0
⇔ ( 3 +1) x 2 + (2 5 + 2 3 )x -3 3 - 2 5 – 1= 0
⇔ ( 3 +1)x2 + 2( 5 + 3 )x -3 3 - 2 5 – 1= 0
∆ ' = b ' 2 – ac= 3 + 5 2 – ( 3 + 1 )( -3 3 - 2 5 – 1)
= 5 + 2 15 +3+9 +2 15 + 3 +3 3 +2 5 + 1
=18 +4 15 +4 3 +2 5
= 1 + 12 + 5 + 2.2 3 + 2 5 + 2.2 3 . 5
= 1 + 2 3 2 + 5 2 + 2.1.2 3 +2.1. 5 + 2.2 5 . 3
= 1 + 2 3 + 5 2 > 0