Question

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

Answer 1

`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}`