Question

2x®-√5x+1=0 it has two distinct groups??​

Answer 1

Answer:

$\blacksquare \: \: \tt \: x = \frac{ \sqrt{5} \pm \sqrt{3 \iota} }{4}$

Step-by-step explanation:

We have equation,

$\blacksquare \: \: \: \tt {2x}^{2} - \sqrt{5} x + 1 = 0. \\ \tt \: where \: as \: \\ \: \: \: \: \: \: \tt \: a = 2 \: \: b = \sqrt{5} \: \: and \: \: c = 1.$

$\blacksquare \tt \: d = {b}^{2} - 4ac. \\ \: \: \: \tt \: = {( \sqrt{5}) }^{2} - 4 \times 1 \times 2. \\ \: \: \: \: \tt = 5 - 8. \\ \: \: \: \tt \: = - 3$

$\blacksquare \tt \: let \: \sqrt{ - 1} \: \: is \: \: \sqrt{ \iota} .$

$\blacksquare \tt \: \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ \: \: \: \: \: \: \: \: \: \tt= \frac{ \sqrt{5} \pm \sqrt{3 \iota} }{4} .$