Q:

what is one root of this equation 2x^2-4x+9=0?A. -2+sqrt14/2iB. -1+sqtr14/2iC. 1+sqrt14/3iD. 1+sqrt14/2iE. 2+sqrt14/2i

Accepted Solution

A:
For this case we have the following quadratic equation:[tex]2x ^ 2-4x + 9 = 0[/tex]The solutions are given by:[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]We have to:[tex]a = 2\\b = -4\\c = 9[/tex]Substituting:[tex]x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (9)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-72}} {4}\\x = \frac {4 \pm \sqrt {-56}} {4}\\x = \frac {4 \pm \sqrt {-1 * 56}} {4}\\x = \frac {4 \pmi \sqrt {2 ^ 2 * 14}} {4}\\x = \frac {4 \pm2i \sqrt {14}} {4}\\x = \frac {2 \pm i\sqrt {14}} {2}[/tex]Answer:[tex]x_ {1} = \frac {2 + i \sqrt {14}} {2}\\x_ {2} = \frac {2-i \sqrt {14}} {2}[/tex]