(i) We have,
2x2 – 1x + 3 = 0
Here, a = 2, b = –7 and c = 3
Now, b2 – = (–7)2 – 4(2) (3)
= 49 – 24 = 25 Since, b2 – 4ac > 0
Therefore, the quadratic equation 2x2 – 1x + 3 = 0 has distinct roots i.e. α and β.
Now,
So,
Therefore, the required roots are 3 and
(ii) We have,
2x2 + x – 4 = 0
Here, a = 2, b = 1 and c = –4
Now, b2 – 4ac = (1)2 –4 (2) (–4)
= 1 + 32 = 33
Since, b2 – 4ac > 0
Therefore, the quadratic equation 2x2 + x – 4 = 0 has distinct roots i.e. α and 946;.
Now,
So,
(iii) We have,
Here,
Now,
Since,
Therefore, the quadratic equation has equal roots i.e.
Now,
Therefore, the requried roots are
(iv) We have,
2x2 + x + 4 = 0
Here, a = 2, b = 1 and c = 4
Now, b2 – 4ac = (1)2 – 4(2) (4)
= 1 – 32 = – 31
Since, b2 – 4ac = 0
Therefore, the quadratic equation 2x2 + x + 4 = 0 has no roots.