Please show how you did it so I can learn :)

Question

asked Nov 9, 2022 188k views

1 Answer

Ivospijker · Answer 1 · 2022-11-14T06:05:59+0000

Answer:

$x_(1) = 3 + \sqrt {6}$

$x_(2) = 3 - \sqrt {6}$

Explanation:

Given the quadratic equation;

x² - 6x + 3 = 0

To find the roots of the quadratic equation, we would use the quadratic formula;

Note: the standard form of a quadratic equation is ax² + bx + c = 0

a = 1, b = -6 and c = 3

The quadratic equation formula is;

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

Substituting into the formula, we have;

$x = \frac {-(-6) \; \pm \sqrt {-6^(2) - 4*1*(3)}}{2*1}$

$x = \frac {6 \pm \sqrt {36 - (12)}}{2}$

$x = \frac {6 \pm \sqrt {36 - 12}}{2}$

$x = \frac {6 \pm \sqrt {24}}{2}$

$x = \frac {6 \pm 2 \sqrt {6}}{2}$

$x_(1) = \frac {6 + 2 \sqrt {6}}{2}$

$x_(1) = \frac {6}{2} + \frac {2 \sqrt {6}}{2}$

$x_(1) = 3 + \sqrt {6}$

Or

$x_(2) = \frac {6 - 2 \sqrt {6}}{2}$

$x_(2) = \frac {6}{2} - \frac {2 \sqrt {6}}{2}$

$x_(2) = 3 - \sqrt {6}$