MATH SOLVE

4 months ago

Q:
# Identify the zeros of the function f(x) = 2x^2 β 4x + 5 using the Quadratic Formula

Accepted Solution

A:

For this case we have that by definition, the roots, or also called zeros, of the quadratic function are those values of x for which the expression is 0.Then, we must find the roots of:[tex]2x ^ 2-4x + 5 = 0[/tex]Where:[tex]a = 2\\b = -4\\c = 5[/tex]We have to:[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}[/tex]Substituting we have:[tex]x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (5)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-40}} {4}\\x = \frac {4 \pm \sqrt {-24}} {4}[/tex]By definition we have to:[tex]i ^ 2 = -1[/tex]So:[tex]x = \frac {4 \pm \sqrt {24i ^ 2}} {4}\\x = \frac {4 \pm i \sqrt {24}} {4}\\x = \frac {4 \pm i \sqrt {2 ^ 2 * 6}} {4}\\x = \frac {4 \pm 2i \sqrt {6}} {4}\\x = \frac {2 \pm i \sqrt {6}} {2}[/tex]Thus, we have two roots:[tex]x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}[/tex]Answer:[tex]x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}[/tex]