Question

Use the quadratic formula to find the solutions to the equation. X^2-3x+1=0

Answer 1

Answer:

x = [-b +- sq root (b^2 - 4 ac)] / 2a

a = 1 b = -3 c = 1

x = [- -3 +- sq root (9 - 4 * 1 * 1)] / 2 * 1

x = [3 +- sq root (5)] / 2

x1 = 1.5 + sq root (5) / 2

x2 = 1.5 - sq root (5) / 2

Step-by-step explanation:

Answer 2

For this case we have the following quadratic equation:

$x ^ 2-3x +1 = 0$

The solution is given by:

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

We have to:

$a = 1\\b = -3\\c = 1$

Substituting:

$x = \frac {- (- 3) \pm \sqrt {(- 3) ^ 2-4 (1) (1)}} {2 (1)}\\x = \frac {3 \pm \sqrt {9-4}} {2}\\x = \frac {3 \pm \sqrt {5}} {2}$

We have two roots:

$x_ {1} = \frac {3 +\sqrt {5}} {2}\\x_ {2} = \frac {3- \sqrt {5}} {2}$

Answer:

$x_ {1} = \frac {3+\sqrt {5}} {2} = 2.62\\x_ {2} = \frac {3- \sqrt {5}} {2} = 0.38$