Question

90%

Answer 1

For this case we must find the solutions of the following quadratic equation:

$x ^ 2-5x-1 = 0$

We solve by means of$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

Where:

$a = 1\\b = -5\\c = -1$

Substituting:

$x = \frac {- (- 5) \pm \sqrt {(- 5) ^ 2-4 (1) (- 1)}} {2 (1)}\\x = \frac {5 \pm \sqrt {25 + 4}} {2}\\x = \frac {5\pm\sqrt {29}} {2}$

Finally, the roots are:

$x_ {1} = \frac {5+ \sqrt {29}} {2}\\x_ {2} = \frac {5- \sqrt {29}} {2}$

Answer:

$x_ {1} = \frac {5+ \sqrt {29}} {2}\\x_ {2} = \frac {5- \sqrt {29}} {2}$