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:

$2x ^ 2-4x + 5 = 0$

Where:

$a = 2\\b = -4\\c = 5$

We have to: $x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Substituting we have:

$x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (5)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-40}} {4}\\x = \frac {4 \pm \sqrt {-24}} {4}$

By definition we have to:

$i ^ 2 = -1$

So:

$x = \frac {4 \pm \sqrt {24i ^ 2}} {4}\\x = \frac {4 \pm i \sqrt {24}} {4}\\x = \frac {4 \pm i \sqrt {2 ^ 2 * 6}} {4}\\x = \frac {4 \pm 2i \sqrt {6}} {4}\\x = \frac {2 \pm i \sqrt {6}} {2}$

Thus, we have two roots:

$x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}$

Answer:

$x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}$

3 0

3 years ago

Please help me find the answer fo this!!

Pepsi [2]

Answer:

÷3==533740£; ok8iejeoeoeo

7 0

3 years ago