Question

Find the roots of the equation 3 X square + 6 X + 1 = 0 by method of completing perfect square please please please help.

Answer 1

$3 {x}^{2} + 6x + 1 = 0 \\ 3 {x}^{2} + 6x = - 1 \\ ( { \sqrt{3}x) }^{2} + 2( \sqrt{3} x) \sqrt{3} + ( { \sqrt{3} )}^{2} = - 1 + 3 \\ ( { \sqrt{3} x+ \sqrt{3} ) }^{2} = 2 \\ = ( { \sqrt{3} x+ \sqrt{3} ) }^{2} - ( { \sqrt{2} )}^{2} \\ use \: identity \: {a}^{2} - {b}^{2} = (a + b)(a - b) \\ = ( \sqrt{3}x + \sqrt{3} + \sqrt{2} )( \sqrt{3}x + \sqrt{3} - \sqrt{2} ) \\ roots \: are \\ ( \sqrt{3}x + \sqrt{3} + \sqrt{2}) = 0 \\ \sqrt{3} x = - ( \sqrt{3} + \sqrt{2}) \\ x = \frac{ - ( \sqrt{3} + \sqrt{2} ) }{ \sqrt{3} } \\ ( \sqrt{3}x + \sqrt{3} - \sqrt{2} ) = 0 \\ \\ x = \frac{ - \sqrt{3} + \sqrt{2} }{ \sqrt{3} }$

Answer 2

given equation

3x^2 + 6x + 1 =0

3x^2 + 6x = - 1

x^2 + 2x = - 1 /3

third term = (1/2 × 2)^2

third term = (1)^2

third term = 1

adding 1 to both sides

x^2 + 2x + 1 = - 1 /3 + 1

( x + 1 ) ^ 2 = 2/3

taking square root of both sides

x + 1 = +/- √2/3

x = -1 +/- √2/3

x = -1 + √2/3 or x = -1 - √2/3