Question

Find the roots of 3x² - 6x +2 = 0
by complete square method.​

Answer 1

ANSWER:

Divide above eq with

x^2-2u+2/3=0

Do half of the middle term

x^2-u+2/3=0

adding both side by (2/3)^2

and final anwer is

u=2-√2/3

Answer 2

Answer:

$x = \frac{3 + \sqrt{3} }{3} or \frac{3 - \sqrt{3} }{3}$

Step-by-step explanation:

$3 {x}^{2} - 6x + 2 = 0$

$( ({ \sqrt{3}x })^{2} - 2 \times \sqrt{3} \times \sqrt{3}x + ({ \sqrt{3} })^{2}) - ({ \sqrt{3} })^{2} + 2 = 0$

$( { \sqrt{3}x - \sqrt{3} ) }^{2} - 3 + 2 = 0$

$({ \sqrt{3}x - \sqrt{3}) }^{2} = 1$

$\sqrt{3}x - \sqrt{3} = \binom{ + }{ - } 1$

$\sqrt{3} x = \sqrt{3} + \binom{ + }{ - }1$

$x = \frac{ \sqrt{3 } \binom{ + }{ - } 1 }{ \sqrt{3} }$

$x = \frac{ \sqrt{3} + 1 }{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{3 + \sqrt{3} } {3} or \frac{3 - \sqrt{3} }{3}$