Question

x²–6x+3=0 by completing the square

Answer 1

Step-by-step explanation:

x^2 - 6x + 3 = 0

=> x^2 - 6x = - 3

=> x^2 - 6x + 3^2 = - 3 + 3^2

=> ( x - 3)^2 = - 3 +9

=> ( x - 3)^2 = 6

=> x - 3 = plus minus sqrt 6

=> x = 3 plus minus sqrt6

Alpha = 3 + sqrt6

Beta = 3 - sqrt6

Answer 2

$\boxed{ \underline{ \red\ast\tt{ \: Solution \: \: \red\ast}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\Rightarrow\footnotesize\tt\red{By \: Completing \: Square \: Method}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\Rightarrow \footnotesize \tt{x ^{2} - 6x+ 3 = 0}$

$\Rightarrow \small \tt{(x - \frac{6}{2})^{2} - (\frac{6}{2} )^{2} + 3 = 0}$

$\Rightarrow \small \tt{(x - \cancel \frac{6}{2})^{2} - (\cancel \frac{6}{2} )^{2} + 3 = 0}$

$\Rightarrow \footnotesize \tt{(x - 3)^{2} - (3 )^{2} + 3 = 0}$

$\Rightarrow \footnotesize \tt{(x - 3)^{2} - 9 + 3 = 0}$

$\Rightarrow \footnotesize \tt{(x - 3)^{2} - 6= 0}$

$\Rightarrow \footnotesize \tt{(x - 3)^{2} = 6}$

$\Rightarrow \footnotesize \tt{x - 3 = \sqrt{6} }$

$\Rightarrow \footnotesize \tt{x = 3 \: \: \pm \: \sqrt{6} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\boxed{ \underline{ \Rightarrow\footnotesize \tt \red{x = 3 + \sqrt{6} }}}$

$\boxed{ \underline{ \Rightarrow\footnotesize \tt \red{x = 3 - \sqrt{6} }}}$